 Welcome back lecture 48. I think today is day 58 actually of a 71-day semester so we are Rounding the corner looking at the finish line Before we go into new material. It was brought to my attention That I had a mistake on the last Problem we went through in class Probably multiple mistakes. This is the only one that was actually brought to my attention so let's go back to that problem and You'll see that if my brain was fully engaged As opposed to about a third engaged We wouldn't have made that mistake actually we I wouldn't have made that mistake so we did I think this was the last problem we did last class We were Trying to find the interval of convergence and the radius of convergence, so here was the original problem This was actually written later We were trying to determine that it was centered around negative two Because that was our x minus a it was actually x minus negative two We used the ratio test. This is not the mistake in fact on this page is not the mistake But this is what we did We got this value for our limit One-third absolute value of x plus two We made sure that was less than one in order for this thing to converge because that's what we're looking for Interval of convergence radio radius of convergence so just convert this absolute value inequality into this algebraic statement and We got our interval To be kind of initially from negative five to one It was centered at negative two and the radius of convergence was three We still had a little more work to do and that is to check the end points and that last one is where my error occurred So we checked out x equals negative five by putting negative five in for x We got negative three to the end Which is negative one to the end times three to the end left a three in the denominator So we decided that was an alternating series We thought it looked like it was in trouble and in fact it did not pass the first test of the alternating series test So that one's okay Actually the conclusion here is okay, but how we got there is not okay So this is where the error occurred Apologize for that x equals one So we put in one for x one plus two is three to the end so we end up with the n over three This is where I apparently was not fully engaged mentally because n over three as n approaches infinity is Not one-third okay, so if you were thinking that go ahead and stop me in that erroneous train of thought that I'm on and n over three as n approaches infinity Clearly is not zero, but it's not one-third as n gets larger the numerator gets larger. This thing gets infinitely large Still the same conclusion so no different answer, but it's not one-third That would be infinitely large. I'm sorry for that. Thank you. I don't know who pointed that out Ben She went that out to me Thank you. Yes, you had negative three to the end and you split it up negative one to the end three to the end When you multiply those together wouldn't it be negative three to the two n no why? Because if you're gathering things together in that type of scenario where you have Let's say two cubed Times five cubed is that the same thing as two times five? Cube you don't do anything to the exponent you bring out that Commonality everything's being multiplied and these things are individual Individually being raised to the power, but you're not going to do anything to the power when you group them together. It's still So negative one to the end and three to the end is negative one times three to the end or negative three to the end so that Well, I get So x squared times x squared in this type of thing would be Because you're really trying to factor out the thing being squared. It'd be x times x Being squared that's the same Line of reasoning and I think all those are correct But if I do make mistakes, please correct me. I need to especially leave it up there when it's the last thing we do Sorry about that. All right We will meet all four days this week Friday is a What they call spring holiday? What we used to call good Friday, but we will not meet that day, but we will meet all Four of the normal days, so it will be a four-day week, but we need all the days to this week to do what we need to do All right, let's keep going. I think we get to the really good stuff Still coming up in this chapter, but this is a good lead into Taylor and Maclaurin series Which is coming up and there's some pretty valuable stuff pretty I think fun stuff in their functions as power series and then we'll See what kinds of power series we can come up with and what are the coefficients look like how can we use the functions and Higher-order derivatives to get coefficients. That's later, but right now. It's just kind of a power series Approach so we're going to have a function. Here's the first one. We'll kind of use this as a Not only as a model, but also it resurrects things that we've already done in this chapter So I think the first time we encountered this we talked about it in terms of an infinite geometric series and This is of the form a over 1 minus r So if this were the sum of Terms in an infinite geometric series the first term would be one and the ratio would be X so we multiply by x as we go so we have this power series in X it is infinite in nature so if The absolute value of the ratio is less than one then this thing converges and in fact it converges to one over one minus x So we can write that as in sigma notation in fact any Of course, it depends where you start in so let's actually do a couple of these So if we have the first term times the ratio to the n minus one, I think that's the first way It's presented in the book We don't really want the first term to Have the ratio in it. We don't want it to be present in the first term So if that's the case then we would want to start this at one So the first term generated would have the ratio to the one minus one or to the zero So it's technically not present but we could write that as a r to the end as long as we started in at Zero and we've been doing some of that lately. So that's probably actually done more in the latter part of chapter 8 than it is in the earlier sections of chapter 8 is starting in at zero I think it in a lot of ways makes more sense to start in at one Then one generates your first term two generates the second term and so on Well, if that's the case here first term is one Times the ratio which is x And if we use this format right here where we're starting it at zero Then we would want to start In itself at zero the first term doesn't have x in it. It's just one So if we just have x to the end In starting at zero and going to infinity. This is the expanded version And this is what it would be as a function. So when it says a function as a power series There's our function for this particular problem our function of x and That power series converges If the absolute value of x is less than one So there's the interval of convergence the radius of it's centered at x equals zero and the radius of convergence is one So let's use that particular model Where this Thing in the numerator when we're done with it is going to represent the first term of our infinite geometric series This thing that is being subtracted from one is Going to be the ratio and really the only thing we have to do as we convert some other function into this form is We have to make sure that this term is One if it's not one we've got to do something to it to convert it to one Once we've got that term to be one It doesn't matter if we're adding something or subtracting something we can figure out what the ratio is It doesn't really matter what's in the numerator because that's indicative of the first term So let's say our function is one over one plus X cubed and the directions are to write it as a power series So the first thing we have to do is if that is not one is to make it one as long as what we do is legal It is one so we're in business We can convert that in fact in this problem. That's probably the next thing and any time you have addition that's convertible to subtraction by Subtracting a negative so is it hot in here? See if we can prop the door open and get a little bit of outside air Moving it may be one of those days where we're kind of transitioning from hot to cold and We don't have any air running. Of course that just wait a couple hours, right? It's supposed to be 30 degrees tonight. That's what I heard Wonderful weather this time of year Go out and breathe all that Yellow-green pollen that just wonderful see it rolling down the street in the rain All right, so the first term Once we have it converted into this form this thing is the first term so our first term is one it looks like our ratio is Negative X cubed so if we were to write out the expanded form It's kind of hard to say what kind of answer we're expected to give When it just says write it as a power series so we could let's do both let's write it in an expanded form Let's also write it in the closed form the sigma notation So if we were writing it out the first term is one the ratio is negative x cubed. So what's the next term? one times negative x cubed Multiply by negative x cubed again. What's the next term? Okay Multiply by negative x cubed so it is alternating in fact We can kind of make it have that look of an alternating series But there it is an expanded form. So that is this function as a power series Under what conditions would this particular power series be convergent Thing is ultimately Go ahead and leave that open. Thanks Kelly We want the ratio which is that an absolute value to be less than one So technically I guess when x cubed is less than one and when would x cubed be less than one Okay negative one to one so the same interval of convergence that we had when the ratio was just x It's going to be the same one for this particular Power series. Let's go ahead and write it using sigma notation First term is one that doesn't really need to be written The ratio is negative x cubed That works doesn't see if when in is zero we generate the first term When in is one do we generate the second term when in is two That term seems to work What are some other forms so just in case you come up with an answer that maybe doesn't exactly match this And we'll just leave that one out because it's kind of Doesn't really serve much of a purpose. So that has the look of an alternating series How about x cubed to the end how else could that be written? something raised to a power itself raised to a power x to the Three in and let's say we want to start this thing at one instead of at zero How do we do that? What about this in right here? in minus one or in plus one some way we just need to make sure it doesn't start with a negative, right? It's got to start with a positive So if you want to use in minus one that'll work in plus one will also work Just have to start it the way we want it to start start with a positive How about the power of x right we want an in minus one right there where there was an in Let's test it if in is one first term is negative one to the zero which is positive X to the what? zero The first term is one. Let's try n equals two negative one to the one which is negative which we want it to be negative if In is two two minus one is one x to the third and that is what we want the second term to be so that seems to work There's a lot of different versions of answers even if yours doesn't match what you see in the back of the book or In case you are submitting an answer to WebAssign, and it's not liking it You've got some other choices that are a hundred percent equivalent Okay, let's find a power series got to work a little bit harder at this one two over three plus x Okay, so again kind of keeping in mind. This is where we're headed Eventually what's left in the numerator is going to be a what is subtracted from one is going to be the ratio and Again the key is getting a one in that position, and there are a lot of ways of accomplishing that There are a lot of things we could factor a two out in front and just not have the two in the numerator But I'm just going to leave that alone So this three that's here. Let's divide it by three, but we can't divide that by three Without dividing everything else in the numerator and denominator by three. So that's almost there. What else? Okay, one minus negative x over three So now we're ready to write it as a power series in expanded form if that's what's desired first term is two-thirds Ratio is There's first term. There's the ratio. So what is the value of the next term of this power series? Two nights x next term Okay, we know it's going to be with the ratio of being negative. We know we're going to alternate signs So we just need to multiply this by x over three right and alternate the sign. So what do we get? And then we've got the pattern going right Alternate the signs Multiply the numerator by x and the denominator by three and we're in business So if it's an expanded form that'll work using sigma notation first term is two-thirds Ratio is negative x over three Okay, we want to start it at zero. We can start our scrolled variable at zero Negative x over three to the n let that thing run all the way to infinity I guess that's not true all the time, but it's true for the values for which the power series converges Where does this converge? Does that work? So we want the ratio To be less than one in absolute value we can drop the negative sign the absolute value of the negative so we want That so we want x over three smaller than one greater than negative one Think you said it multiply through by three the larger interval of convergence larger radius of convergence Let's get some other versions here could I bring that two-thirds out in front Nothing variable in there and we could separate out the alternating part so negative one to the end So that would work and we could change the scrolled variable to have it started n equals one and doctor that up a little bit That's enough I think For that one. All right. Here's a function First time we've had something. That's not just a number in the numerator x cubed Again, here's where we're headed a over one minus r Okay, did you think through what you would want to do? Divide by four that'll probably do it on it. What's first term? Nice delightful what? X cubed over four Ratio is positive or negative x over four Okay, it's whatever is being subtracted from one the first term x cubed over four See what happens with this we might get a variety of answers here the first term is X cubed over four And the ratio is x over four to the end Can anything there be brought out front? Okay, we could bring a one-fourth out front and we've got x cubed That four gets raised to higher and higher power. So we better leave that one alone So x over four to the end is x to the n over four to the end any advantage to writing it that way Okay, we put our x's together Here's x cubed here's x to the end. They have the same base. They're being multiplied. We can add the exponents so x to the n plus three Over Four to the end and let's check out a couple and see if that is in fact what we want So it any let's keep the one-fourth out in front of everything So at n equals zero we've got in plus three which is zero plus three x to the third over Four to the zero is that right? Is that what we want for the first term? That's right at n equals one x to the one plus three x to the fourth over four to the one Does that seem to be right? Equivalent to what we wrote earlier. There's first term X to the fourth over sixteen Seems to be working right what? advantage is this form Over our original form. It's a little bit simpler to substitute into but it to be honest with you It's not enough different to make a big deal over instead of Putting in your end value and adding three to every exponent. It's basically taking care of right there Interval of convergence Not convergent for all values When is this convergent? Okay negative four to four does that work? Because we would want the absolute value of x over four to be smaller than one in absolute value You'll be able to it's kind of hard to get pictures of this now But as we move into later sections of chapter eight Taylor and McLaurin series, you'll be able to get a picture of what the Interval of convergence is all about what it looks like and if you go beyond that interval of convergence Kind of what happens to the graph? Inside that interval and outside the interval so we'll be able to get it eventually a visual on that All right power series can be differentiated And can also be integrated so let's make that the Kind of the task here for the rest of the class and we'll get in numerous Examples of that so let's start with some generic power series where we've got a Just a number C sub zero as the first term Coefficient of the n equal one term will be C1 We'll either have an x or an x minus a Generically we can probably do better by saying so there's our power series So using sigma notation it seems like each term has a C Subscripted So let's call it C sub n and we're going to start in at zero if that's the case because our first term has a C sub zero We're going to have powers of x minus a And if we start not that that's necessarily where we have to end let's make sure it's right If in it is zero does that start where we want this to start? When in is zero we'd have x minus a to the zero and we don't want an x minus a right here When in is one we've got a C sub one x minus a to the one So it looks like the subscript and the power are in agreement right and we want that to happen from our first line Let's take the derivative term by term That we'll have also another way to do that which is what this is This little section is all about C sub zero is a number its derivative is zero What's the derivative of C sub one x minus a? If we distributed this we'd have C C sub one times x and C sub one times a a is a number So the derivative of C sub one times a would be zero derivative of C sub one x is C sub one is that okay? What's derivative of the next term? Derivative of the next term And isn't that pattern going to be the same the rest of the way If that pattern is present the same way we had a similar pattern, but slightly different pattern in the original function Here was our pattern For the original function Let's see if we can write that pattern out Using sigma notation and is that pattern one that we can derive without using the expanded version How could we write this? We're going to have C sub something Okay, you want to have an in here Now that's not really present here, so we really don't pick that up till here, right? So we don't have the same type of agreement we had up here the Subscripted C and the power of x minus a were the same Here the subscripted C and the power of x minus a are different C sub 2 x minus a to the 1 C sub 3 x minus a to the 2 so it looks like it's one behind so that seems to be true and in I Guess we can kind of figure out where we want to start this In equals one that correct is that going to work? If in is one And I don't we'll need to justify that to why we think it needs to start at one when the original started at zero When in is one the first term ought to be one C sub one X minus a to the zero is that our first term in the derivative it's correct, isn't it when in is two We ought to have a two C sub two X minus a to the two minus one is that the next term Seems to be giving us what we want So if that's our original function and this seems to be working for the derivative Do we really have to do a term by term? the fourth term Isn't that going to be sixty three on the top one, but only four see three on the bottom one? All right, let's make sure that's good. That's a good thing to do So if we did the derivative here, we would want three times two would be six Maybe we need to doctor that up a little bit the next term up here would be sorry wrong line The next term up here would be C four X minus a to the four. You know we're still okay, aren't we? Yeah, I was thinking like you right I was thinking maybe we needed factorial when you made that comment But I think we're okay because we're not gathering any Prior coefficients along the way. We've just got the four from the power. I think we're okay So I'm gonna write the two things that I have circled and get the other clutter out of the way So here's our f of x and here's our f prime of x So can we do something with this? To get this just take the derivative, right? Isn't that our kind of our shortcut? Term by term so we can do it in the closed version now the only other thing that we would probably have to justify is Why did this one start at zero and this one starts at one? You're exactly right. I was wondering if somebody was going to say that you can start this one at zero, right? You want to because if in or zero You'd have zero times C sub zero x minus a to the negative first I don't know that we want to get into x minus a to the negative first, but it really doesn't matter because it's zero anyway, right? So zero in a sense generates the first term, but the first term that actually is Nonzero starts when in is one so that's another that's guess. That's a good way of thinking of it You could start it at zero that doesn't generate any terms at all generate zero which is the derivative of Our constant term from f of x if you want to start it with the actual terms that actually get Generated we can start in at one another way to justify it is don't we lose a term? Right Because our first term was constant We're going to have it seemingly one fewer term the derivative of that constant is zero So it doesn't show up so you really do in a sense have the choice of starting it in n equals zero But the first term you really want to Generate that's non zero starts when n equals one So if you took our just our regular power rule or exponent rule You take the exponent times the coefficient that's already there times the same thing to one degree last that's exactly what this is and We got this by looking at the term by term pattern that was there So it's Valid so you can take the derivative in its closed form Let's go to the integral So what we want to do is integrate the function We can do a term by term integration Why don't we take the other path this time since we took the term by term path the first time What would it look like if we integrate it? So if we started right here? And now we want to integrate Not term by term, but just this thing right here. What would it look like? We'll see the n plus c sub in we bring that Coefficient along for the ride don't we do that in differentiation and integration? We'd have x minus a to instead of to the end. We'd have it to the We're integrating now in plus one divided by n plus one Is that work Should we we can check it with a couple of them and we would want to start in at zero Yes, because we'd want to that c sub zero term when we integrate it We want it to be there and in fact it will now be more than just C sub zero So when in a zero C sub zero x minus a to the First all over zero plus one which is one that's the first term Should be don't necessarily agree with that We might ought to spend a few seconds on that but let's go on to see what the next term See if it's kind of getting us what we want when in is one What do we get c sub one? x minus a to the two Over two let's write out what f of x was it was c sub zero plus C sub one x minus a C sub two So if we integrated that guy right there, do we get this? That's got a little bit of a It works If you integrated c sub zero with respect to x wouldn't you just really get c sub zero x Right We've got c sub zero x minus a is that okay Well, there's the c sub zero x. We've got a little extra baggage c sub zero times minus a what is that? That's a constant and aren't we integrating and gonna when we're done We're gonna have to put a big old plus c at the end of this thing Anyway, isn't that part of the plus c it's kind of absorbed the nice thing about that is that we do end up with Powers of x minus a all the way down the line So that one is a little tricky to get by the rest of my think are pretty easy C sub one x minus a to the two over two That's what we've got here n equals two C sub two x minus a to the third over three So it does in fact give us what we want gives us a little extra here let's put a K value in front of that so that there is the possibility of some arbitrary constant any time we do an indefinite integral How are we going to know what that is or how might we find k in? One problem or another how do we typically find a c or a k in an integral problem? They have to give us something extra. What's typical of the extra f right a point f of zero equals five Plug in zero for x and it's supposed to kick out five Something like that, so we'd have to have some additional information to find k, but we will do problems For which we will find this k value and we'll be able to Decide what the actual solution is and know what the k value is for that particular problem So you can differentiate power series by did basically differentiating the argument You can integrate power series by integrating the argument of the power series in the sigma notation We can start this. I don't know if we'll finish it So we want to express that as a power series is that function and if we get this far We can stop and we'll pick up from this point tomorrow is this function in any way shape or form related to The one that we started class with today and did several examples modeled after this Okay, it's the whole thing squared Keep in mind what we just finished doing differentiating and integrating power series What's the derivative of that? What's the derivative of one over one minus x well, that's the same thing as That right what's the derivative of one minus x to the negative first negative one Minus x to the negative second times the derivative of what's inside Hey, how about that? It's one over one minus x Squared and that's what we want So is the one that is given to us in some way shape or form related to Related to in this case meaning is it the derivative of or is it the integral of Something that we've already done kind of in a basic case Yes, this one right here is in fact the derivative of One over one minus x well if we want a power series for this guy and we already have a power series for this guy What do we do to it? Take the derivative of it. Okay, that's actually is a pretty good place to stop So we can pick up with how we go about doing that tomorrow