WEBVTT mathematics/calculus-ii/murray
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Hi, this is Will Murray for educator.com and we are here to talk about power series.
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A power series is a series of the form, it has a coefficient c < font size="-6" > n < /font > ,
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Then it has a power of either x^n or x-a^n.
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The game with power series is you are trying to plug in different values of x and see what values of x.
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When you plug in a value of x, you get a series just of constants.
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Then the question is which values of x make that thing converge.
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If you try out some examples, we will work on some examples soon.
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You start to notice a pattern which is that you always have a center value at a.
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Now you notice that if you plug in x=a here, then you get a-a^n, so you just get a series of 0's, that always converges.
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It always converges at x=a.
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It turns out that it always converges in some radius around that center.
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So, there is always some radius r around that center.
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And the series always converges absolutely between those values.
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That is a+r, and this is a-r.
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It always converges absolutely in that region.
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That region is absolute convergence.
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It is always divergent outside that region.
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Then at the two end points, a-r and a+r, for those values of x, it is really unpredictable what the series does.
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Sometimes it will diverge at both of them, it could converge at both of them.
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It could converge at one and diverge at the other one.
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There is no way to predict it ahead of time, just every particular example, you have to check each one of those endpoints separately.
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To recap here, it always converges between a-r and a+r.
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Another way of saying that is that x-a, in absolute value, is less than r.
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Because it is saying x is within r units of a.
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r is called the radius of convergence, it can happen that the radius is 0 or infinite.
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We will see some examples of that.
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Then the interval, what you can do is you can look at the interval from a-r to a+r.
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That interval is called the interval of convergence.
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That interval might include both endpoints or not include both endpoints, or it might include one of them and exclude the other one.
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There are all of these different possibilities about whether the endpoints are included.
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That interval, including whichever endpoint makes it converge is called the interval of convergence.
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That is a lot of definitions to start with.
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It will be easier once we look at some examples.
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Here is an example, we want to look at the interval of convergence for the power series x^n/2^n × n.
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The trick with power series is you almost always want to use the ratio test.
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You almost always want to start out using the ratio test.
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We will start out using the ratio test.
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Remember the ratio test says you look at a < font size="-6" > n < /font > +1/a < font size="-6" > n < /font > in absolute value.
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That in this case is x^n+1/2^n+1 × n+1/x^n/2^n × n.
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All in absolute value.
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I will flip the denominator up and multiply it by the numerator.
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I am going to organize the terms with the terms that look like them,
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We get x^n+1/x^n.
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2^n/2^n+1, and n/n+1.
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I cannot get rid of the absolute values completely because the x could be positive or negative.
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This simplifies down into the absolute value of x.
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n/n+1, when you take the limit as n goes to infinity, that just goes to 1.
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Then 2^n/2^n+1 just gives you a 2.
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Remember, the ratio test says it converges absolutely whenever that limit is less than 1.
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We look at abs(x)/2 < 1, and that is the same as saying the abs(x) < 2.
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So, x is between -2 and 2.
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Those are the values of l that give you a value less than 1.
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Those values of x make this series absolutely convergent.
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If x < -2, or x > 2, we know that the ratio will be greater than 1.
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The series will be divergent.
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Let me fill in what we have learned so far.
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From -2 to 2, we know that it is absolutely convergent.
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If it is less than -2, we know it is divergent,
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Or if x is bigger than 2, we know it is divergent.
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The question we have not answered yet is what happens at the endpoints.
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A big rule here is you cannot use the ratio test at the endpoints.
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The reason is the endpoints are exactly where the ratio test gives you 1.
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The ratio test, when it gives you 1, tells you nothing.
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You have got to use some other test at the endpoints.
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Let us check the endpoints.
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Without using the ratio test because I know if I use the ratio test, I am just going to get 1.
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Let us check these separately.
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If x=-2, then our series turns into -2^n/2^n × n, which simplifies down into -1^n/n.
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That is a series that we have investigated before.
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This converges by the alternating series test.
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That is the alternating harmonic series, and the alternating series test applies to it.
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x=2, gives us the series of 2^n/2^n × n, which gives us the series the sum of 1/n.
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That is the harmonic series, or if you like, that is a p series with p=1 and we know that diverges.
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We have seen that one before.
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That is the harmonic series.
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You can also think of it as a p series with p=1.
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The important thing there is that 1 is less than or equal to 1.
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Anything < or = 1 makes it diverge.
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So x=-2 makes it converge, x=2 makes it diverge.
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So, the interval of convergence is everything between -2 and 2.
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And we include -2 because it makes it converge.
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We exclude 2 because it makes it diverge.
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Another way of writing that in interval notation is to say the interval from -2 to 2,
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You put square brackets on the -2, that means it is included,
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Round brackets on the 2, that means it is excluded.
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That is your final answer there.
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Let me recap here.
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The important way to approach power series, most of the time you want to start out with the ratio test.
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At least 95 times out of 100, you want to start out with the ratio test.
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You work through the ratio test and you get your limit.
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You set that less than 1, because remember less than 1 is what the ratio test looks for to tell you the series converges.
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That tells you the basic interval except it does not tell you the endpoints.
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It tells you that it is absolutely convergent between those points, divergent outside those points.
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Then you check those two endpoints separately.
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At first you do not know what the endpoints do, and remember, you cannot use the ratio test for the endpoints, because the ratio test will give you 1.
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When the ratio test gives you 1, you get no information and you have to try something else -- you have to plug them in individually.
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x=-2, it turns out that that one works,
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x=2, it turns out that it makes it diverge.
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When we are writing the interval we include x=-2, and exclude x=2, and if we write that in interval notation, we put square brackets on -2, and round brackets on 2.
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Let us try another one.
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We want to find the interval of convergence for the power series of -1^n × x^n/n!.
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Again, we will start with the ratio test.
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We will look at a < font size="-6" > n < /font > +1/a < font size="-6" > n < /font > .
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I know that these -1's will get swept away by the absolute values anyway, so I am not even going to write the -1 terms.
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I am just going to write x^n+1/n+1! divided by the a < font size="-6" > n < /font > term, x^n/n!.
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Still need my absolute values here.
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If I flip my denominator and pull it up, we get x^n+1/x^n × n!/n+1!.
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That simplifies down to x abs(x).
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Now n+1!, remember you can write that as n! × n+1.
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So, the n!'s cancel and we are just left with x/n+1.
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Remember, we take the limit of this as n goes to infinity.
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We are plugging in different values of x here, but whatever value of x we plug in,
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It is a constant. n is the one going to infinity.
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When n goes to infinity, no matter what value of x you start with, this thing goes to 0.
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For all values of x, no matter what value of x you start with.
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This thing goes to 0 because n is the one going to infinity.
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So, l is 0 no matter what you start with.
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The key point here is l < 1, so the ratio test says that no matter what value of x you put in there, it is absolutely convergent.
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For all possible values of x, for all real numbers of x.
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If you want to write that in interval notation, the answer would be (-infinity, infinity).
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I put round brackets there because there is no question of endpoints here because infinity and -infinity are not actual numbers we can plug in there.
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There is no question of plugging in endpoints or considering endpoints to be included or excluded.
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There are no endpoints.
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We say that the interval runs over all real numbers from -infinity to infinity.
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Another way of saying that would be that x going from -infinity to infinity.
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We started here by applying the ratio test.
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We wrote down the ratio, we took the limit as n goes to infinity, because that 0, which < 1, no matter what value of x it is, we get that it converges for all possible values of x.
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Our third example here, we are not given a power series, we have to write a power series ourselves.
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For ln(1-x).
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Then we have to find the interval convergence.
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This one is not so obvious how to start unless you have seen something like this before.
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The key point is to remember that the derivative of ln(1-x),
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Is 1/1-x × derivative of 1/x, so that is -1/1-x.
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1/1-x is exactly the sum of a geometric series.
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That is the sum from n=0 to infinity of x^n, put a negative there.
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That is a geometric series and that is true for abs(x) < 1.
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That is probably easier to think about if you work backwards.
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If you look at this geometric series, that sums up to the first term/1 - the common ratio,
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So that is equal to 1/1-x.
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Armed with that intuition, we can write ln(1-x) as the integral of 1/1-x dx.
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Except there was that negative sign, so I will put that on the outside.
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That is the integral, now 1/1-x we said was this geometric series.
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That is 1+x+x², and so on, dx.
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If you integrate that, you get negative, integral of 1 is x, integral of x is x²/2,
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Integral of x², is x³/3, and so on.
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But this is not an indefinite integral, so we always have to include a constant.
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How do we figure out what the constant should be.
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To find the constant, we are going to plug in x=0 to both sides.
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We get ln(1-0) = -, well if we plug in 0 to a bunch of x terms, we get a bunch of 0's + C.
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So, the constant is ln(1), which is 0.
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So that is nice, the constant disappears.
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We get ln(1-x) is equal to, I will distribute the - sign, -x - x²/2 - x³/3, and so on.
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We can write that as the sum from n=1 to infinity, of -x^n/n.
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Notice that that is what we would have gotten if we had integrated this geometric series directly.
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If we took the integral of x^n, we would have gotten,
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Well, x^n+1/n+1, then shifting the indices from n=0 to n=1,
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Would have converted that into x^n/n.
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That is the power series.
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What we now have to do is find the integral of convergence.
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What we knew before, is that x was between 1 and -1.
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When you take the derivative of integral of a power series, it does not change its radius of convergence.
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We know it still goes from -1 to 1.
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However, it might change what happens at the endpoints.
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So, we still have to check the endpoints of this series.
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Sorry, check the endpoints of the integral, and see whether it converges or diverges at those two endpoints.
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Wee have to check those separately.
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Let us try x=-1 first.
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that would give us the series of -1^n/n,
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That is the alternating harmonic series.
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We know that converges by the alternating series test.
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x=1 gives us the sum of -, well 1^n/n,
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That is the negative of the harmonic series,
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And we have seen several times that that diverges, either by just saying that that is a harmonic series,
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Or by saying it is a p series, with p=1,
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Since 1 < or = to 1, that makes it diverge.
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x=1 makes it diverge, x=-1 makes it converge.
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So, the interval, you could write that as x between -1 and I am including -1 because it made it converge,
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I am not including 1 because it made it diverge.
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In interval notation that is [-1,1),
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With a straight bracket on -1 because it made it converge and a round bracket on 1 because it made it diverge.
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Our answers there, we found the series by taking the derivative of ln(1-x),
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We get something we can convert into a geometric series,
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Then to get back to ln(1-x), we took that geometric series and we integrated it back up to get ln(1-x).
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Now we have a power series for ln(1-x).
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To get the interval, we used the interval for the geometric series, but then because we are taking derivatives and integrating,
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We know that the endpoints, whether or not it converges at the endpoints, that could possibly change.
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We have to check those again, we check -1, it converges.
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1 diverges, and so we reflect those in the answers for the interval.
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We will try some more examples later on.