WEBVTT mathematics/differential-equations/murray
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Hi and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and we are going to be doing a review of powers series today.
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The reason is that later on we are going to be learning how to use power series and Taylor series to solve differential equations.
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Since it might have been a while since you actually studied power series and Taylor series, we are going to work through some of the basic properties today.
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In the next lecture you will be ready to go to use power series and Taylor series to solve differential equations.
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Let us jump right in, here were going to start out with the review of the definition of Taylor series.
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You start with a function f(x)and you take a point x₀ and we find the Taylor series expansion around that point which means the sum, remember the capital letter σ means the sum.
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You are adding these terms up from n=0 infinity of a sub n, x times first times X minus X₀ to the end where the a's of n, these coefficients are fn(x₀)/n factorial.
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Just a couple reminders here remember that this fn what that means is the nth derivative of f and derivative of the function f.
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When n=0 that means the zero derivative which means you're taking the derivative zero times which just means the original function f and when n=0 that is just f₀ itself, n factorial means you multiply the numbers 1 x 2 x 3 x 4 up to n, that is what n factorial means.
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Let me remind you of that 1 x 2 x 3 up to n and by convention we say zero factorial is equal to 1.
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That is something that throws off a lot of my students at first but it makes all the formulas later on work out nicely so it is worth accepting that convention that zero factorial is equal to 1.
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Most common type of Taylor series that we study here is when x₀ is equal to 0 so the the point here x₀ would be equal to 0 and that is called the Maclaurin series.
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Maclaurin series just means a special case of the Taylor series when the x₀ is equal to 0.
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In that case, the Maclaurin series of s of a function is the sum from n=0 to infinity of a sub n(x^n) and the a of n is still the same.
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It is the sum from n=0 to infinity of the nth derivative of f applied to x₀ would be 0 because we are doing the Maclaurin series divided by n factorial x x^n.
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That is the Maclaurin series and that is probably 90% of the examples you study in calculus and differential equations, maybe even more than that or actually involving Maclaurin series which is the special case of the Taylor series.
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There are certain Maclaurin series that it is really worth memorizing, hopefully you memorize these back in your calculus class and I will just give you a little reminder of them now.
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But if you did not memorize back then, it is worth memorizing them at least while you are using series to solve differential equations.
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Just very common functions are e^x sin x cosine x and 1/1 - x.
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The Maclaurin series for e^x is 1+ x + x²/ 2 factorial + x³/3 factorial.
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The general pattern there is that this is the sum from n=0 to infinity of x^n/ n factorial.
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Remember that if n=0 then you get x⁰, you just get 1 and then you get n factorial.
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0 factorial is just 1, this is the n=0 term gives this 1 over here the n=1 term gives the x and then it is a little more obvious in the terms after that and how they correspond to the general sum.
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The next one that you really want to remember and recognize is sin x which is x - x³/3 factorial + x⁵/5 factorial -x⁷ factorial.
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It is a little alternating series and the way we can make it alternate is we put a -1 to the n term.
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That makes it alternate the positive and the negative and the way we get the odd powers there is we do x² n +1.
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As n ranges overall number 01234, 2 n + 1 will range over the odd numbers 1357.
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That is the way of collecting those odd powers and we want the same odd numbers in the denominator.
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We put 2n +1 factorial in the denominator, that is a close form expression for the Maclaurin series for sin x.
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For cos x, it is quite similar except you have the even powers 1 - x²/ 2 factorial plus x⁴/4 factorial - x⁶/6 factorial.
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Again we can now write this summation form for that, the sum from n= 0 to infinity again it is alternating so I'm going to put it somewhere -1^n in there
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And then to get the even powers I'm going to go x^2n/(2n factorial).
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Be a little careful there with the 2n factorial, some people think that (2n factorial) is equal to 2 factorial x n factorial or 2 x n factorial, that is emphatically not true.
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You have to be very careful with the parentheses there 2n factorial, for example of n=10 and then 2n factorial would be 20 factorial which is not the same as 2×10 factorial, you have to be quite careful with the parentheses there.
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The last series that we are going to use very frequently when we are studying differential equations is the Geometric Series 1/1-x.
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That expands into 1+ x + x² + x³, that is the simplest one to write down a formula for, that is just the sum from n= 0 to infinity of x^n.
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If you plug in the n=0 term you will just get that first one there and that is true for the absolute value of x < 1, in other words x between -1 and 1.
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That is not valid when x is outside that range, outside interval, that it is valid when x is inside that interval.
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That brings me to the next topic here which is where did these series converge? What values of x are these series are good for?
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That is something that we study back in Calculus 2, by the way if all of these is feeling more rusty then you can brush off in a single lesson.
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You might want to go back and look at the Calculus 2 lectures Here on www.educator.com.
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We spent several lectures back in Calculus 2 going over Taylor series explicitly so you can get at a more lengthy rundown on all these topics back in Calculus 2.
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But if you remember those pretty well and you just need to brush off a little rust then keep playing along here.
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Any power series has a radius of convergence around x₀ so what that means, let me draw a number line here
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The center of the radius of convergence is always that value x₀ and then what you do is you go a distance, the same distance in either direction and that is the radius r.
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That means the top and the interval you have x₀ + r and at the bottom of the interval you have x₀ - r.
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The idea is that when you are inside that radius the series converges, in this green region here the series converges.
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When you are outside that radius this series diverges, what that means is if you plug in values of x that are outside this radius.
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If you plug those values into the series then the series diverges and this is called the Radius of Convergence.
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A shorthand algebraic way of keeping track of this is to write this as x - x₀ in absolute value is less than r because that will exactly capture this values between x₀ - r and x₀ + r.
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By the way, when I say x nought, nought means 0. So that is just x…saying x nought is just a shorthand way of saying x⁰.
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The big question here is whether it converges at the end points, x= x₀ - r and x=x₀ + r and we do not know that just when we find out the radius.
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The series are quite unpredictable, sometimes it will converge at one end point but not the other one, Sometimes they will converge on both end points, sometimes it will converge at neither one, we do not really know that until we test each one individually.
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There are some extreme cases to this radius, If r=0 that means that the series converges only at x₀ and at nowhere else.
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That is the case where r=0 and that is really a pretty worthless series because if you think about it if you plug in x=x₀, x = x₀ that would mean that x - x₀ is 0 so you are just getting a series of all 0, 0 to the end and that is really a worthless series.
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If you add up a bunch of 0 you will certainly get 0, it certainly converges which you do not really learned very much, r=0 can happen but it is essentially a worthless case.
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The other extreme of that is r=infinity, r = infinity means that the series converges infinitely far in either direction so it converges for all points on either side of x₀, that is a very good case.
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It means you can plug in any value of x you like and the series will converge, that is the best case if the radius is infinite.
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The way you want to find the radius of convergence of a taylor series or have any power series, it is usually to use the ratio test now back in Calculus 2 we learn a whole lot of different task.
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We learned the alternating series test, the integral test, the ratio test, and root tests and all these tests in Calculus 2.
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When you come right down to Taylor Series and Power Series the one you use almost all the time is the ratio test, that is the only one that I'm really going to mention here.
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But if you want some more details on checking when series converge, you might want to go back and check those lectures for Calculus 2.
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The way you check if a series converges is you form this ratio where you look at the nth term so that means if you have a generic term then you would plug in for that generic term.
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Then you would plug in n+1 into that generic term, you go back and replace all the n's by n +1 and you divide them together, that is the ratio.
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That is why it is called a ratio test, you take its absolute value which is nice because it gets rid of any negatives that might be hanging around.
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You take the limit as n goes to infinity and what you generally get is some expression in terms of x and you set that less than 1.
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That will tell you for all those values of x wherever values of x you end up solving for.
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Those are the values for which these series converges and the values of x that make this limit greater than 1 diverge.
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The way this looks on a number line is you have x₀ and then you will have some region within which that limit will be less than 1.
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Outside that region I will put this in red here, that limit will be greater than 1 on either end and so it converges when the limit is less than 1.
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Maybe I will put that in green to emphasise that that is the convergent region, limit here is less than 1 and that is the convergent region.
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Now the big question here is the end points and the ratio test never tells you what happens at the end points x₀ - r and x₀ + r.
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You do not know from the ratio test whether the series converges there and what you have to do is plug both of those values in separately.
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Each one into the series separately and see whether it converges using some test other than the ratio test.
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The ratio test is guaranteed to fail at the end points because it will give you 1 and that is the cut off value so instead you use some other ratio, some other test besides the ratio test that we learned back in Calculus 2.
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Let us see how all this works out, first example here is to identify the following power series as an elementary function.
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We are given the series 1+ 3x² 9/2x⁴ 9/2 x x⁶ 27/8 x x⁸ this is a typical kind of thing you will get at the end of the differential equations problem.
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Where you have a series as your solution and you are trying to recognize that as some function maybe that you have seen before somewhere.
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Let me show you how to think about this, I see 3x² is really nothing clever.
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I can do and that and I see 9x⁴ and what I noticed is that (3x²)².
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I'm going to write it that way and it looks like I'm starting to build up some kind of pattern here, I'm going to write (3²)³ for the next term but (3x²)³.
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That would give me 27 here and all I had was a 9, to balance that out I'm going To divide by 3 in order to balance out the fact that all I had here was 9.
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I'm going to keep going here, I will put (3x²)⁴ but 3⁴ is 81 and since I had a 27 here I have to divide by 3 to keep it balanced, that is 3×8 which is 24.
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This Is 6, I'm noticing that I can write this in terms of powers of 3x², as long as I modify the denominators accordingly.
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I'm going to write this as 1+ 3x² + (3x²)², I look at these numbers in the denominator and I see 2, 6, and 4, that is pretty clearly a factorial pattern so 6 is 3 factorial and 24 is 4 factorial.
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What I recognize here is something that looks like the series that I have memorized for e^x , let me write that down just quickly up here.
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Remember the e^x is 1+ x + x²/2 factorial, 2 factorial is just 2 + x³/3 factorial + x⁴/4 factorial and here 2 factorial is the same as 2.
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I'm going to write that is 2 factorial, I see that I have exactly the same series here except that I got 3x² wherever there was a and x in the series for e^x.
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What this gives me is e(^3X)².
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What I did there was to look at the series that I was given and I noticed that the powers of x are going up by 2, I know that there is something about the power of x².
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I also noticed just for the first 2 terms I had 3 and 3² and so I thought maybe I have some pattern and powers of 3 there.
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That is why I started making patterns of 3x² all the way through here and then I had to put some extra numbers in the denominator just to make that pattern work out.
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When I looked at the numbers that I ended up with the denominator, I noticed that I got this factorial pattern 2, 6, 24 is 2 factorial, 3 factorial.
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Once I got it into that form I remembered my series for e^x and I noticed That I have the same series where x has been replaced by 3x²
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That is how I know that this whole thing is equal to 3x².
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Next example we are going to find the Maclaurin series for the function f(x)=natural log of 1-x.
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There is a very seductive mistake that can be made here that calculus students often make, they use the generic formula for Maclaurin Series, which is to write down the sum for n=0 to infinity of f^n of 0/ n factorial x x - x₀^n.
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But x₀ is 0, it is just x^n, I will start writing down derivatives and I will look for a pattern.
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That can often be the longest and slowest, most difficult way to find a Maclaurin series and the reason is sometimes writing down the 3rd and 4th derivative of a function gets very messy.
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You are very liable to make mistakes and it is very hard to spot the power so instead of going from that basic formula and just writing down derivatives,
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I want to introduce another method which is to try to relate this function to one of the ones from which we already have memorized series.
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What I know is that natural log of x, if I take the derivative of natural log of x, dy/dx of natural log of 1-x.
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What I get is 1/1-x and in the chain rule x -1 so the point of that is that 1/1-x, I have a memorized prefabricated known Maclaurin series.
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I'm going to use that Maclaurin series that I already know for 1/1-x and use it to build a series for natural Log of 1-x.
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Let us see how that works out, formally natural log of 1-x I know I can get that by starting with 1/1-x and taking its integral of one little issue here which is this -1 right here.
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I will put in negative outside here and I will throw on a dx, I'm going to take the integral of 1/1-x and that is going to give me natural log of 1-x along with a negative sign.
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The point here is that I know a series for 1/1-x so my series for 1/1-x that was the geometric series 1+ x + x² + x³ and so on.
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When we had that back at the beginning of this lecture that was something we figured out long ago in our calculus lectures the series for natural log of 1-x, this is just like a polynomial and it is very easy to integrate.
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If I integrate this, I have that negative and the integral of 1 is x and the integral of x is x²/2, the integral of x² is x³/3 and so on.
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Whenever you do an integral you have to add an arbitrary constant so I will put in a constant on the outside.
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I'm going to put it on the left here and so this is c - x, I need to distribute that negative sign so it's minus -x²/2, -x³/3 and so on.
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That is supposed to be equal to natural log one of 1-x and I do not know what that constant is yet so in order to figure out what that constant is I'm going to plug in a value of x.
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The easiest value of x for me to plug in is x=0 and I'm going to plug that in to both sides, I will plug in x=0 to both sides and on the left the natural log of n=0 , so that is natural log of 1.
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On the right, I will get c- a bunch of 0 and natural log of 1 i know that that's equal 0, I will get c=0.
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I know what my c is and i can go back and fill that in and natural log of 1-x= -x - x ²/2 and I dropped out the c to now because i figured out it was equal to 0- x³/3 and so on.
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I can write that in σ form as the sum from n= it looks like it has to be 1 to infinity because I'm starting at n=1 here.
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x^n/n and there is still a negative sign in there so that is my series for natural log of 1-x my maclaurin series for natural log of 1-x that might be one the you remember doing in calculus 2.
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Just to recap there what happened was I had to find the maclaurin series and this is my generic formula for maclaurin series for a function.
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But even though it is tempting to use that formula and start writing down derivatives it is easier to relate the series and function to one of those known functions that we already have a series for.
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That is what I try to do over here is I try to remember that the derivative of natural log of 1-x is the same as 1/1-x except for a negative sign.
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I know the maclaurin series for 1/1-x, I am excluding that fact very heavily so I set up natural log of 1-x is the integral of 1-x and a negative sign convert that 1/1-x.
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This is using the fact that I already memorized the series for 1/1-x convert that into its series form .
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That is just a polynomial so it is very easy to integrate I just integrate it term by term and 1 gives me x, the x gives me x²/2, x² gives me x³/3 and so on.
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I also have an arbitrary constant so that's my series for natural log of 1-x except that I have not figured out what that constant is.
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So to find that constant I plug in x= 0 to both side so 0 there and then 0 for all this x's and on the left that gives me natural log of 1which is 0.
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On the right, that just gives me c, I figure out that c=0.
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When I plug that back in I get my series natural log of 1-x is -x-x²/2 - x³/3 and so on.
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I can write that in σ form like this and that is my maclaurin series for natural log of 1-x.
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In example 3, we are going to find the interval of convergence for that maclaurin series that we just figured out.
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Let me remind you what that series was, we figured out that natural log of 1-x is equal to the sum from n=1 to infinity of x^n/n.
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Not n factorial just x^n/n and it's negative so the way we are going to find the interval of convergence is we are going to look at the ratio test.
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Remember the ratio test says you look at the n +1 term and divide that by the n term of the n term and the denominator first, because the easier part that is -x^n/n.
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The n+1 term means you replace all the n's by n +1, so -x^n +1/n + 1, that is the limit that we had to simplify.
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Some good things happen right away which is that because we have absolute values all the negative signs go away.
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I do not have to worry about negative sign, I will write that down that it is killed by the absolute value and now we are dividing a fraction by a fraction so I'm going to flip the bottom fraction up.
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I have n/x^n x x^n +1, just flipping that bottom fraction up to the top, it is still an absolute value and I'm going to reorganize things a little bit and I'm going to sort all the x's together.
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x^n +1/x^n x n/n + 1 and we are supposed to be taking the limit of that and I'm going to separate these two terms x^n +1 /x^n.
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That just simplifies down to x and the limit as n goes to infinity of n/n +1, you can divide top and bottom by the highest power.
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That is n here so this limit gives me 1 in the numerator and then n/n is 1+1/n and the 1/n goes to 0 so that whole thing goes to 1.
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This whole thing gives me just the absolute value of x after I take the limit.
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The idea with the ratio test is that limit is supposed to be less than one for the series to converge, what that tells me is that this is where the series converges.
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Absolute value of x less than 1 that means x is between -1 and 1 and it diverges outside of that region, do not diverge if x is less than -1, x bigger than 1.
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The only thing that ratio test does not tell us and it never tells us is whether or not the series converges at the end points.
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We always have to check the endpoint separately and you can not use the ratio test to check the endpoints ever.
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We are going to plug x= -1 and x=1 into that series separately and we can not use the ratio test anymore.
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We plug in x=1 into that series right here then we will get the sum from n= 1 to infinity of x^n will just be 1/n, that is the negative of 1+ 1/2 + 1/3 + 1/4 and so on.
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This is not something we discussed yet in these lectures on differential equations but it is something we talked a lot in the lectures on calculus two and I'm hoping that it something you remember.
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If not you might want to go back and check at the lectures on calculus two. but this is the harmonic series that we studied in calculus 2.
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It is an example of a p series if that rings a bell for you and what we learned in calculus 2 is that, that diverges.
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The harmonic series diverges, that means that x=1 is not at the interval of convergence because it makes the series diverge.
00:33:06.000 --> 00:33:24.000
Let us check the other endpoint that means we are going to plug in x = -1 in here into the series, -1 or n=1 to the infinity x -1^n.
00:33:24.000 --> 00:33:42.000
For the harmonic series, when n=1 here we get -1¹ x -1, we get +1 that -1^n essentially gives us the same series except it is going to be alternating, that is going to be 1-1/2-1/3-1/4.
00:33:42.000 --> 00:33:52.000
This is something we have not studied here in differential equations yet but it is something we studied back in calculus 2.
00:33:52.000 --> 00:34:10.000
We figured out that this is not alternating harmonic series which does converge by what we call at that time the alternating series test.
00:34:10.000 --> 00:34:25.000
Go back and look at the lecture in calculus 2 on alternating series and you will see this is actually one of the examples we studied back then.
00:34:25.000 --> 00:34:33.000
We figured out that x= -1 does make it converge, x =1 makes it diverge.
00:34:33.000 --> 00:34:46.000
Let me summarize this.
00:34:46.000 --> 00:35:10.000
x does converge 1x= -1 the series does converge and when x=1 it does not converge that is why I'm putting less than equal to, for x = -1 and less than for x =1, that is my interval of convergence and sometimes you write this in interval notation just different notation for the same thing.
00:35:10.000 --> 00:35:23.000
We are going to write the interval from -1 to 1and we are going to include the end point at -1, we are going to put square brackets there and we are going to exclude the end point at 1, we will put a round brackets there.
00:35:23.000 --> 00:35:37.000
Such as different notation for the exact same thing square brackets on one side to show that endpoint is included and round brackets on the other side to show that end point is excluded.
00:35:37.000 --> 00:35:57.000
By the way, we could have possibly predicted this, I will say that this makes sense if you think about the original function that we studied here which is natural log 1-x.
00:35:57.000 --> 00:36:08.000
Let us see what happens if you plug in natural log of 1-x, if you plug in the x=1.
00:36:08.000 --> 00:36:20.000
If you plug in x=1, that is natural log of 1-1 so it is natural log of 0 and if you remember the graph of natural log, it looks like this.
00:36:20.000 --> 00:36:27.000
At 0, it kinds of blows up to negative infinity, it is undefined.
00:36:27.000 --> 00:36:42.000
It approaches negative infinity so it makes sense that the series would not converge at x=1 because it is trying to approximate a function that is going to negative infinity.
00:36:42.000 --> 00:36:45.000
We got our interval here, let me recap what we did.
00:36:45.000 --> 00:36:56.000
This was the series that we derived back in example 2, that is all coming from example 2, you might want go back and check that if you have not just watched it.
00:36:56.000 --> 00:37:06.000
The way we figured out the interval of convergence was we applied the ratio test, which means you look at the n + first term divided by the nth term.
00:37:06.000 --> 00:37:19.000
We took our formula and we plug in n +1 for n and then we just left with n + n in place of n and immediately the negative signs disappeared because of the absolute values.
00:37:19.000 --> 00:37:33.000
We sorted our fractions here and the we separated out the x's on one side and the n's on the other and the x^n+1/x^nand just cancels down to x the end of n +1.
00:37:33.000 --> 00:37:41.000
We take the limit of that and it turns into 1, that whole terms since we are multiplying it that just all drops out turns into 1.
00:37:41.000 --> 00:37:53.000
We are left with just the absolute value of x and remember the ratio test tells us that whatever you get in that limit, it converges when that limit is less than 1.
00:37:53.000 --> 00:37:59.000
The absolute value of x has to be less than 1 for it to converge.
00:37:59.000 --> 00:38:11.000
We know it converges from -1 to 1, diverges outside that interval we don't know what happens at the end points remember the ratio test never tells you what happens at the end points.
00:38:11.000 --> 00:38:25.000
You got to check each one separately, we plug that x =1 into this series right here, plug in x=1 here and we got something that we remember from calculus is the harmonic series.
00:38:25.000 --> 00:38:39.000
The negative of the harmonic series which diverges, that was something we learned back in calculus 2 and if we plug in x=-1 we get the alternating harmonic series.
00:38:39.000 --> 00:38:54.000
We also learned in calculus two that converges by the test we learned from calculus two was the alternating series test so what that tells us is that it converges for x =-1 and it diverges for x=1.
00:38:54.000 --> 00:39:10.000
The series converges from -1 is included up to 1 excluded and you can write that interval notation n by putting square brackets on the [-1] and round brackets on the (1), that is our interval.
00:39:10.000 --> 00:39:24.000
Of course we checked because natural log 1-x the original function that we started here if you try to plug-in x=1 then you will be trying to take natural log of 0 which does not exist.
00:39:24.000 --> 00:39:33.000
That is a nice check on the fact that our series did not converge at x =1, let us move on.
00:39:33.000 --> 00:39:49.000
In example 4, we are going to use power series to solve the integral of e(^x)²dx and this is really starting to illustrate the usefulness of power series.
00:39:49.000 --> 00:39:57.000
Because this is an integral that you would not have been able to solve by the standard integration techniques in calculus 2.
00:39:57.000 --> 00:40:09.000
The only way you can solve this is by some kind of approximation or by using taylor series techniques so we need to find the power series for e(^x)².
00:40:09.000 --> 00:40:20.000
Now that is something that if you start to write derivatives it will get very messy, very quickly you will get tangled up in the chain rule, product rule, and the power rule.
00:40:20.000 --> 00:40:36.000
It will all get very quickly involved and get very messy, instead the way to write down a power series for e(^x)² is to remember what your power series is you put to the x.
00:40:36.000 --> 00:40:43.000
I hope you got that memorized, if not i recommend that you memorize it at least while you are going to be using series to solve problems.
00:40:43.000 --> 00:40:51.000
But here it is, it is 1+ x + x² 2 factorial + x³/3 factorial and so on.
00:40:51.000 --> 00:40:57.000
That was for e^x but then you can get the series for e(^x)² just by making a little substitution.
00:40:57.000 --> 00:41:08.000
Whenever I see an x, I'm going to change it to x² and I get 1+ x² + x(²)², well that is x⁴.
00:41:08.000 --> 00:41:20.000
It still over 2 factorial, that does not change, plus x(²)³, that is x⁶/3 factorial and so on.
00:41:20.000 --> 00:41:34.000
What I really want to do is integrate both sides here, so I put my integral sign here and squeeze in a little dx where I can, dx.
00:41:34.000 --> 00:41:44.000
The key point here is that now that I have converted the function into what is essentially a polynomial that goes on forever but we can think of it as a polynomial.
00:41:44.000 --> 00:41:53.000
I can integrate this term by term very easily the integral of 1 is just x, integral of x is x³/3.
00:41:53.000 --> 00:42:05.000
Sorry the integral of x² is x³/3, the integral of x⁴/2 is x⁵/5 x 2 factorial.
00:42:05.000 --> 00:42:16.000
The integral of x⁶/3 factorial is x⁷/7 x 3 factorial and it will keep going along that pattern.
00:42:16.000 --> 00:42:42.000
Whenever I do an indefinite integral I always have to add an arbitrary constant and I'm going to add that at the beginning so that is my integral e(^x)²/2, I'm sorry e(^x)²dx.
00:42:42.000 --> 00:42:58.000
You could also use the summation form to write this, an alternate solution would be to say that e^x is the sum of x^n/n factorial.
00:42:58.000 --> 00:43:09.000
e(^x>)² is just the sum of x(²)^n which is x^2n/n factorial.
00:43:09.000 --> 00:43:22.000
When you integrate that, when you take the integral of both sides, these sums all go from n=0 to infinity.
00:43:22.000 --> 00:43:41.000
The integral of x^2n his x^2n+1/ 2n + 1 x n factorial.
00:43:41.000 --> 00:44:03.000
This is another way to arrive at the same result if you expanded out that series you get that same thing that we got over here except I forgot to include my c, so I will tack that on.
00:44:03.000 --> 00:44:13.000
We found our integral of e(^x)² in terms of a series which is something you never could have done in calculus 2 using the basic integration techniques there.
00:44:13.000 --> 00:44:24.000
Let me remind you how we did that, I remembered my series for e^x and that is much easier than trying to find the series for e(^x)² directly by writing down derivatives.
00:44:24.000 --> 00:44:31.000
I just substituted in the x, i changed it to x squared, I made that substitution on both sides.
00:44:31.000 --> 00:44:39.000
x(²)² gave me x⁴, x(²)³gave me x⁶.
00:44:39.000 --> 00:44:49.000
We made that substitution on both sides that converted it into essentially a polynomial, certainly in terms of integrating it i can think of it as a polynomial.
00:44:49.000 --> 00:45:04.000
When I integrate both sides, my 1 turns into x, x² goes to x²/3, x⁴/2 factorial goes to x⁵/5 x 2 factorial and so on, we just integrate it term by term.
00:45:04.000 --> 00:45:20.000
I tack on the arbitrary constant and the other way to think about that is to remember that e^x we have a nice σ formation for the series and e^2x, sorry e(^x)².
00:45:20.000 --> 00:45:43.000
We just substitute in x² and so we get x^2n in this series and the integral of that is just x^2n+ 1÷ 2n + 1, that is a way of getting a series of a σ summation form for the same thing.
00:45:43.000 --> 00:45:51.000
On our last example here we got y(x) is the sum from n=0 of a sub n, x^n.
00:45:51.000 --> 00:45:58.000
We are being given a generic power series and we want to find the power series expressions for y′ of x and y″ of x.
00:45:58.000 --> 00:46:14.000
We are going to shift the indices of summation so that they start at n=0, this is actually an example that is designed to warm you up for using power series to solve differential equations which is what we are going to be using in the next lecture.
00:46:14.000 --> 00:46:27.000
I really encourage you focus on how this example works because essentially the start of every problem in the next lecture is going to be exactly what we are doing in this problem.
00:46:27.000 --> 00:46:43.000
It is really a key problem here but let us see how it works out if y(x) is given to us, this generic power series, that is the guess we will make to solve differential equations in the next lecture the Y′(x).
00:46:43.000 --> 00:47:01.000
We are going to take the derivative there n=0 to infinity of, now I will put in a sub n because that is a constant but the derivative of x^n is n x x^n -1.
00:47:01.000 --> 00:47:19.000
We get this new series but you will notice that the n=0 term is 0 and that is because of this factor of n right here so that n=0 term of that series really doesn't change anything.
00:47:19.000 --> 00:47:38.000
In fact i can leave that off and just start the series at n=1 so we can omit that n=0 term and I can rewrite this series exact same thing except for n=1.
00:47:38.000 --> 00:47:51.000
That is because instead of adding up the n=0, 1, 2, 3 terms I noticed that the n=0 term really is not doing anything, I can just add up the n=1, 2, 3, 4 terms.
00:47:51.000 --> 00:48:06.000
I'm writing the same thing here (an)x < n^-1 and it is something that is shifting the indices of summation so that they start at n=0.
00:48:06.000 --> 00:48:37.000
I want to show you what I mean by that if I expand out this series, I'm going to start out at n=1 and I will get a sub 1, x⁰ is just 1, I will not put anything there, that was from n=1 term, the n=2 term gives me 2 a sub 2 x x¹ , n=3 term gives me 3(a3) x x² and so on.
00:48:37.000 --> 00:48:56.000
I'm going to forget the old version of the σ form and just look at these terms and if I want to write it in σ form now and I want to write so that there is an x^n there.
00:48:56.000 --> 00:49:20.000
What I noticed is these numbers on the coefficients are 1 higher than the power of x, here I got 3 a sub 3 x² so the coefficients are 1 power lower, what I'm going to do is I'm going to put n+1.
00:49:20.000 --> 00:49:42.000
I think I did leave myself quite enough space here, let me rewrite that, n + 1, a sub n + 1, x^n and if I do that that all make sense if i start at n=0.
00:49:42.000 --> 00:49:59.000
If you think about that if I plug in n=0 here, I will get 1 x a1 x x⁰ which would give me just exactly the a1 term here.
00:49:59.000 --> 00:50:13.000
The n=1 term would give me 2 a sub 2x, the n=2 term would give me 3 a sub 3² and I'm going to end up getting the exact same series and I wrote above.
00:50:13.000 --> 00:50:25.000
I'm just using different index of notation to keep track of that and let us look at what effectively happened here from when I expanded it out and wrote it in a new form.
00:50:25.000 --> 00:50:58.000
If you look at this n=1 term, the index went down by 1 so we lowered the index by 1, because of 1 down from 1 to 0 and then all the n's in the formula, all these n's got raised by 1.
00:50:58.000 --> 00:51:07.000
That n turned into an n+1, the n and the subscript term turned into n+1, and the n+1 turned into an n.
00:51:07.000 --> 00:51:37.000
To fix that you raise the n's in the formula by 1, that is going to be a really useful trick for us to use in the differential equations lecture next time.
00:51:37.000 --> 00:51:47.000
You really want to keep this in mind if you lower the indices by one then you raise the n's in the formula by 1 and it works the other way too.
00:51:47.000 --> 00:51:59.000
If you raise the indices by one then you lower the n's in the formula by 1 and it is a way of adjusting the powers so that we can make different series be compatible with each other.
00:51:59.000 --> 00:52:10.000
Let us practice that by going one more steps here by finding y″(x) so y″(x) , that is the second derivative.
00:52:10.000 --> 00:52:21.000
Again I will start with n=0 to infinity now I got an x^n -1, while I still have an n.
00:52:21.000 --> 00:52:35.000
n^x-1 if i take its derivative will give me n-1, there is still an a sub n, x^n-2 , that is by taking the derivative of this expression right here.
00:52:35.000 --> 00:52:53.000
I noticed that the n=0 term because of this n right here will be 0 and the n=1 term, let me write that down the n=0 term is 0.
00:52:53.000 --> 00:53:13.000
Because of that n-1, the n=1 term is also 0, I can drop those two terms out and just start at n=2 to infinity of nx n-1 x a sub n/ x^n -2.
00:53:13.000 --> 00:53:21.000
I'm not shifting the indices of summation yet, I'm just dropping off the first couple of terms because I noticed that they are going to be 0.
00:53:21.000 --> 00:53:26.000
I'm not doing this business over here where I'm shifting indices that is what I'm going to do next.
00:53:26.000 --> 00:53:34.000
I looked at this and I see that x^n -2 right there, I want to raise that up and make it x^n.
00:53:34.000 --> 00:53:57.000
What I'm going to do is I'm going to raise the n's in the formula by two in order to get that n -2 up to an n.
00:53:57.000 --> 00:54:25.000
To make that work I have to lower the index by 2, by the same principle, I will not bother writing out the series this time, it is the same principle as before, so that n=2 in the index, I'm going to drop that down to n=0 and I'm going to raise all of these n's by 2.
00:54:25.000 --> 00:54:41.000
I have n turned into n +2, n-1 raise it by 2 and gets it to n+1, a sub n becomes a sub n+2, x^n-2 becomes x^n.
00:54:41.000 --> 00:55:04.000
Let me box what I have figured out, your Y′ is n= 0 to infinity of n+ 1, (an) + 1x^n and Y″ is n= 0 to infinity and +2 x n +1 x (an) +2 x x^n.
00:55:04.000 --> 00:55:12.000
These are really important formulas, it is worth studying them very carefully and making sure you understand them or at least to remember them.
00:55:12.000 --> 00:55:19.000
Because we are going to be using them very heavily next time when we use series to solve differential equations.
00:55:19.000 --> 00:55:33.000
Let me recap how we obtained those formulas, we started with y′ of x I took the derivative of this so the derivative of x^n is nx^n-1 .
00:55:33.000 --> 00:55:44.000
I noticed at the first term that n would just be 0 anyways, I dropped it out and I started at n=1, instead at n=0 and dropping off the n=0 term.
00:55:44.000 --> 00:55:59.000
I did this shifting index trick where I lowered that one down to 0 and then I raised all these n's and the way we can justify that if you want to is to expand out both this series and that series.
00:55:59.000 --> 00:56:07.000
If you expand out either one you will see that you will get this series either way, those are really the same thing.
00:56:07.000 --> 00:56:21.000
That was my formula for y″, I took another derivative of x^n -1 and I got n-1, x^n-2.
00:56:21.000 --> 00:56:34.000
I noticed that since I have an n and an n-1, the n=0 and the n=1 term are both going to be 0 so I drop those out and I can start my index at n=2.
00:56:34.000 --> 00:56:46.000
That was not shifting the index yet that was just dropping off a couple of 0 terms and I'm going to use the same principle where I will lower the index by 2, that is why that 2 became a 0.
00:56:46.000 --> 00:57:03.000
Then I raise each of these n's by 2, so that n became n +2, that n -1 became n +1, that little n became an n + 2, and that n - 2 became an n.
00:57:03.000 --> 00:57:13.000
If you do not trust that, if that seems like magic, just expand out each one of these series, expand out this one and expand out this one right up the first few terms of each one.
00:57:13.000 --> 00:57:28.000
You will see that you will get the same series either way and this is going to be absolutely crucial for us when we use differential equations to solve our new series to solve differential equations at the topic of the next lecture.
00:57:28.000 --> 00:57:32.000
And that concludes our review of power series for this lecture.
00:57:32.000 --> 00:57:38.000
You have been watching the differential equations lectures on www.educator.com. My name is Will Murray, thanks for watching