 Welcome back lecture 50 at some point in time today in class. We will start and Hopefully make good progress on Taylor series, but I understand we have a couple of web assigned questions Web assigned is not due tonight due to our night before I forget to do this. This is not open yet But on the 13th starting April 13th the class evaluation Things will actually be available So that'll go for that'll get all the classes that you're registered for but And then I know that's going to be on the cable broadcast and eventually on the DVDs for this class, but Cable TV students also will be doing evaluations and then future semesters that are taking this class on DVD through Distance Ed will also have an opportunity to do evaluations. So it's appropriate to To know that and to think about those but those become active on April 13th. All right web assigned questions What one was that? five Okay, our tangent is the same as inverse tangent Let's see what we already know about that So I think yesterday we did just inverse tangent of x and we decided that was the integral of 1 over 1 plus x squared integrated with respect to x So that's not anything new. We've seen that we actually saw that the first time in Calc 1 where we took the derivative of the inverse tangent and we got that Rational function, so now we're anti differentiating that and getting our way back to inverse tangent We also Came up with a power series for inverse tangent yesterday, and what did that look like? X to the To in yeah now as that we integrated it we had to integrate a power series to get to our final answer So do we want to enter to n plus one to n plus one? over to n plus one and We were starting at let's see if we can This particular series that we integrated is that so the first term is one and the ratio is negative x squared So that's what we got for that one right and then we integrated that. Let's go ahead and write what that was That was negative one to the in x to the To in Now as that start where we needed to start the first term We need for the first term to be positive Negative one to the zero so we're okay x to the two in n equals zero so that seems to be correct So we integrated that and that's where we're trying to come up with this right? So do we still want the negative we're not going to change the sign right because we integrate anything? This negative is still negative anything that's positive. It's still positive x to the two in plus one Over two in plus one So that's an inverse tangent Is that correct? We might use this so let's write it what what's it look like so the first term is positive and it is x to the one over one so it's x and then we've got a negative x to the Three over three That's what it should look like right Integral of one is x integral of negative x squared is that x to the fifth over five X to the seventh over seven and so on so that should be inverse tangent of x So can we use that to come up with the inverse tangent of x over five? And if so, how can we use that? Negative x squared with Negative x okay, we could we could go all the way back here so if this is inverse tangent of x and we want inverse tangent of x over five Where there's an x here. We could put in an x over five. I Don't know that that's going to be it entirely is that going to work? I mean it will but you just have to go back through And to put up the lives right we've got to I think we've got to do a little bit of correction here because This is x the thing that's being squared is x and this thing is the derivative of x The thing that's being squared here is x over five this is not the derivative of x over five What is the derivative of x over five one one-fifth? So we can put a one-fifth in here as long as we do what multiply by five Does that work? I think that we've kind of mirrored what this statement says for inverse tangent of x We've now replaced x with x over five and dx with derivative of x over five What we decided we actually kind of Mentioned this but then we experienced it firsthand because we tried to use that formula to get the inverse tangent of two And it didn't work and we said it was outside the interval of convergence What is the interval of convergence for this particular function on inverse tangent of x? negative one to one any guesses what the new interval of convergence will be Okay, I've heard two answers. I think they both sound kind of valid as initial stabs at it negative five to five or Negative one-fifth to one-fifth Let's see where the Let's get to the interval of convergence in a minute. If this is the route. We're going to go I think there's a better route, but if this is a route. We're going to go We would have to write out the power series for this right and Integrate it and that should be the power series for inverse tangent not awful, but could we possibly do this? If this is inverse tangent of x and we want the inverse tangent of x over five Could we put an x over five everywhere? There's an x That sound like something we do with series like this In fact, let's do it both ways and see what happens So that might be a power series and it just says write the power series for and find the interval of convergence The radius of convergence That would be one of the things I would want to try but since we have it set up the other way before we leave this page again There's our radius of the thing that we had to integrate and We decided that the radius had to be Less than one because this is technically an infinite geometric series So negative x squared less than one so x squared Less than one and that's where our interval comes from right negative one to one comes from that statement right there Now if we do that Replacement in there then this becomes negative x over five right squared is the radius so if we Work with this so a is the first is the numerator. So this is the first term This is going to be the radius So the first term is and I know we're not to the final answer yet, but let's See if we can get to that same final answer the first term of this Series would be one the next term would be what? X over five squared right When we had our radius as negative x squared We just multiplied by negative x squared as we proceed now our radius is negative x over five the quantity squared So we want to do that It's alternating because the ratio is negative. So the next term is positive and it is what x over five To the fourth and then x over five to the sixth Now if that is That quantity that I've circled which I hope it is Then we would want to integrate that is that correct? So if we integrated that Then we would get what what would we get here if we integrated this? As if math class isn't bad enough now. We have these dental sounds right? Give us a nice recollections of the dentist's office. That's these nice pleasant things going on in this class Okay, that's x squared over 25 so the integral of that would be We've still got the 25 right got a 1 over 25. We're bringing along and then we've got it x cubed over 3 that work and here we've got x to the fourth over 25 squared 625 So it'd be x to the five over five Right, let's compare and contrast our solutions. We did that this way. Well here. We've got x cubed over That's not the same as it Now wait a minute what we do with our five Yeah, take out that three and extra three for the inverse tangent of x over five We've got a little extra baggage out here five did I leave that out? That's gonna make a That's not gonna make it worse Yes, so we don't really need to do all this so the question is not what the series is It's the radius of convergence. Oh, okay. Yeah the radius is negative x over five Squared so we want that to be less than one so we can drop the negative and then we're just going to have I mean you can Probably see what's going to happen as you spread that out algebraically that we'd have to multiply through by five, right? Yeah, absolutely. Thank you Right. We've got x squared. We've got the five squared also so 25 So we multiply through the inequality by 25 there we go, so our first two stabs added neither one of them hit, right? because it was negative x squared between One and negative one now we've added a denominator of five, but it's really being squared so the denominators really 25 but That's x squared negative five to five so we're back where we were Centered in zero. Well, we kind of knew that because there's no x minus a So it looks like Now I still am not I'm gonna have to do a little bit more work We should get the same answer both ways whether we sub into the existing formula for inverse tangent or we derive it ourselves We've got what appears to be two different results for inverse tangent of x over five Well we Don't really do that this way you say x is we're going to square it So could it be as low as negative five and still be less than 25? You really don't pay attention to that because you're not going to have something squared Certainly, it's greater than negative 25. In fact, it's going to be greater than zero But when you square it what could it be at the other end it could be as low as negative five could be as high as five So it's not really taking the square root taking the square root We're not really technically doing that Katie in the second one it is the same as the first one if you multiply the denominator by five okay, but Five cubed Is it? I mean the first one I can see if we divide that by five we were matching here This is really x cubed over five cubed Which is 125 and then we're dividing that actually multiplying it by a third Yeah, because you divide that by five and then you divide it by three just like you did in the first one It's five cubed by five. Okay. It is. Yes. All right. So if we took that so that's all I need to justify now this one is Really one-fifth of that one right? this thing right here If what we said is correct is one-fifth of this one, right? That should be the case here, too. So that should be one-fifth of This one. So we're off by a factor of five Get this out in front Integral you multiplied by five on the outside instead of dividing like when we made the correction Yeah, I think we're I think we're okay. Let's see if this works So we took the inverse tangent of x and Everywhere in the formula we replaced x with x over five and we came up with this one and then we compared that To the one that we kind of developed ourselves How did we go about developing that ourselves? We went all the way back forget the integral sign we went all the way back to the a over one minus r format We developed our own series then we integrated each term of that series but So we really did this Multiply by one-fifth on the like extended series or whatever unless you're multiplying by a fifth on the integral like from one above it And you're multiplying by five Right, we've got this is what I think you're saying so we've got a kind of an extra one-fifth here And that's the one-fifth that we're kind of not addressing here That we've got Or off by a factor of five That's there. That's not on the other side. So you have to divide by five on both sides. Oh Yeah Because the five is you put the five on that side without you need to take it away from the other side So you divide each side by five then you have one-fifth on the other side. Yeah, but didn't I have to have that five? I couldn't come up with that one-fifth Yeah, but tell I think he's gonna put it in there and compensated with a multiplication by five You have to fix the other side like the extended series side like you have the five out in front of the One over one minus negative X over five squared. Yeah, but you don't have it in front of the other one So then you need to divide both sides by five to make them equal Well, that's what made them equal is putting that one good from the five right. That's what made it equal to the right side So you don't You don't need to add a file, right? No, I mean we're off by a factor of five I've just got to justify how all the sudden we come in here now and divide everything by five Which that's gonna make them equal. I know what's gonna make them equal, but how how do we? Justify that division by five from what we did here Which one of those answers is right? Well Because there's two of them which one are we trying to make the other one be like I think that this is the cleanest way of getting there. So we that we know that one. Yeah, I think this one's right Okay, here's the denominator in the other side So you need five in the top of the other one so you would have put one fifth in front of the integral sign Because it's the denominator of the denominator Okay, here's what I think I'll do it and then I'll spend some more time with it and then we'll revisit it The way we have stopped here How does that compare with what we really need in order to integrate this thing isn't it one-fifth of what we need Aren't we missing a one-fifth? Is that correct This one-fifth that we kind of put in here and corrected for so in a sense. Don't we have one-fifth of what we need? That's to me. That's the thing that's working right now So this is really not in a sense all that we need it's one-fifth of what we need So that means this side is also one-fifth of what we need But you divide it by five not multiply it by five Well, right. That's one-fifth multiply by one-fifth. So I need a one-fifth here I can't multiply this side by one-fifth without also multiplying the other side by one-fifth. I Think that's going to justify it, but I need to put some more thought into that But I think we have the radius of convergence And it should be the simpler substitution should also work here, so we should get x over five We should get x over Excuse me x cubed over 375 I think is right So I think that's where it is. I'll let me put some more thought into that and we'll start class with that Tomorrow we have a fifth of what we need on this side, so we should have a fifth of what we need on this side Great any more questions was there another one? That's okay, I mean that It's a battle sometimes. I mean I don't necessarily think that every problem needs to be done in you know 92 seconds or just scrap it. I mean sometimes it takes more than that Yeah, yeah, I was going to listen to it first before we decided And you need to find this form of it k plus the sum of B to the N B sub in no beat to the end to the end Okay, the end times X to the CN plus D C in plus D plus D over all that e to the E N plus F Not e to the end. Just e times Capital E capital E. Yeah an unknown constant. Yeah times in Down here. Yes, I'll eat way too many letters Well, we're going to integrate a series so we're going to have a constant so that's taken care of Let's take this guy off to the side And put that in the form of a over 1 minus r So it's already in that form almost We've got the one which is what we have to have is that work so the first term is X and the ratio is Negative X to the fifth so what negative X to the sixth Plus X to the 11th minus X to the 16th and we want to integrate that Well, what is this? Let's figure out what this thing is first The first term is X. Let's go ahead and write that down. Normally. It's been a one So we just kind of delete it and the ratio is negative X to the fifth So there's our ratio to the nth power. Let's start in at zero and Let it run to infinity Does that work that seem to be describing? This thing Okay, I think we're okay is it to the five in plus one well we eventually want to integrate We don't we're not there yet We're just trying to get a power series group together some like terms and then we'll do the integral of this power series So negative one to the end. Let's separate that out. So we know it's alternating Then we're going to have an X to the fifth to the end which is X to the Five in right and then we've got another X over here So X to the five in Plus one does that work for this one? Everybody content with that and that is not what we want, but that is This part right here. So now if we want to integrate that it's in here. So we've got X to the one and Then we've got a negative one to the end and we've also got an X to the five in so these two We put together, right? Yeah So we want to integrate This thing and then we can check it because we have the expanded version up here We know when we integrate this we better get what? X squared over two And when we integrate this we better get X to the seventh over seven And X to the 12th over 12 and so on so we can check what we have here Not going to change the sign So I guess B from this format is going to be negative one, right? B to the end will be negative one to the end X to the five in plus one Add one to the exponent, right? Now let's see if it gives us what we want so for the n equals zero term We better get X squared over two. Is that work? That works. Yeah for the n equals one term. Yeah, we better get X to the seventh over seven I think it's going to work. I think that's what we want So that seems to be what we want, but there's the cn plus d so five n plus two E n plus f same thing and it looks like B is negative one And then we integrated so we've got also some arbitrary constant that work All right, let's get started with Taylor series A sub category of Taylor series They're called the chlorine series. So the larger category so let's start with a Power series kind of generic power series So we're going to have some coefficient the coefficient is going to change in fact. That's where this That's the emphasis of the Taylor series is dealing with these coefficients in the pattern by which they change So what it looks like expanded this function that can be written as a power series is When in a zero we get C sub zero X minus a to the zero, so I'm not going to write that down C sub one X minus a to the one C sub two X minus a squared I'm going to write several terms because we're going to lose some along the way because we're going to be taking a derivative So that's kind of where we're starting section eight point seven is these things called power series what they're closed form looks like in terms of sigma notation and we want to More or less isolate. How do we decide what? Coefficient's Affect each term as we move out to the right and what what pattern would actually describe them So let's take the derivative of this expanded version Well the derivative of Well, let me go back step We're going to do that. We need to get C sub zero first Before we lose it because when we take the derivative, we're going to lose it So on this side every if everywhere there's an X if I replace it with an A So I'm going to take the instead of the f of X. I'm going to take the f of a I should do the same thing on the right side everywhere. There's an X. I should replace it with an A When we take every X value that's on the right side and replace it with an A What's going to happen? Every term that's X minus a Will now be a minus a a minus a all those are going to drop out. Is that correct? So the only term that's going to remain is C sub zero So we'll just kind of set that aside. We're going to come back to that So before we take the derivative if we put a in for X It just so happens that the f of a is C sub zero All right now. Let's take the derivative So the derivative of C sub zero. It's a constant. So it's gone. What's the derivative of? C sub one times X minus a Just see some one What's the derivative of C sub two X minus a? Square quantity squared that work to the one derivative of C sub three X minus a the quantity cubed And one more for C sub four cubed You know just do that. So the pattern has been established Just like we did up here before we took the derivative everywhere. There was an X. We put it in a let's do the same thing here With the derivative so everywhere. There's an X. We're going to replace it with an a what happens to all the terms that have X minus a and They're gone. They're all zero. So f prime of a is C1 We'll frame that we'll come back to that We don't it's kind of hard to see the pattern until you get a couple steps beyond where we are now So we took the derivative We got C1 Plus two C2 X minus a to the first All right now. We're taking the derivative of C1. What's derivative of C1? It's gone. What's derivative of two C2 X minus a? What's derivative of three C3 X minus a the quantity squared Okay, six, so I'm just going to leave that as two times the three that's already there X minus a to the first right What's derivative of four C4 X minus a the quantity cubed? Four times the three so three times the four that's already there X minus a square and I told you we're going to lose some terms so We're going to trust that the pattern is the same from this point forward Now we've got the second derivative of the function. What is the second derivative evaluated at a? Every term that has an X minus a in it on the right side is going to drop out because those are all zero Let's do one more Derivative of what we have immediately above this derivative of two C2. That's gone What's derivative of this? two times three C3 is that it that's the derivative of this term Derivative of the next term Okay, so two times three times four and again, we're out of terms. We've been losing terms, but What's the third derivative evaluated at X equals a? Hey, how about that? All right, so let's gather up what we do have I probably would be good to get one more Actually, let's get one more the fourth derivative So we're going to this is a constant. We're going to lose that derivative of this is that right The next term is going to have an X minus a in it In fact, every other term beyond that's going to have an X minus a in it So the fourth derivative at a Now let's gather up the things that we have framed along the way and see if we can get a pattern So we have the f of a Is c sub zero the f prime of a is c1 Second derivative at a third derivative at a in fourth derivative So you can probably predict the pattern Factorial was mentioned so two times three times four. We can certainly multiply by one and it's not going to change that We can multiply this by one multiply this by one And multiply this by one so it's not going to change anything to multiply it by one if I want to solve this For C sub four because we're trying to figure out some pattern for C sub in What is C for equal to fourth derivative? at a divided by for factorial Let's solve it the one above it for C sub three it would be third derivative at a divided by three factorial So it looks like we have a pattern. Let's just make sure the pattern works all the way back C2 looks like it ought to be second derivative at a over two factorial that seems to work C1 should be first derivative at a over one Factorial that's this one gets a little tricky, but we want the pattern to be the same So I guess let's just see what the pattern is when this subscripted letter Number for C is for don't we want fourth derivative and for factorial when it's three We want third derivative and three factorial when it's two We want second derivative two factorial when it's one you want first derivative one factorial So what should it look like for zero? zero factorial Thankfully by definition zero factorial is one so that's going to work And it'd be the zeroth derivative I guess it means we haven't taken the derivative So it does work the pattern that we're seeing at work all the way back does work in the Inequal zero case So in general for all these different coefficients CZ C sub zero C sub one C sub two C 11 C sub 187 they should all be the same they should be well somebody tell me what to write down for C sub in The nth derivative At a in fact oil Seems to be the pattern so that's kind of the first step in Coming up with what's called a Taylor polynomial. So let's go back to our power series the power series for f of x We wrote earlier today and also earlier in this class as That and now we know what we can plug in for C sub in in terms of the original function Might seem like way too much to memorize, but there's some nice Values that are in common. So for C sub in we're going to replace it with the nth derivative At a over in factorial And then we've still got the x minus a to the n So here's what makes this kind of complicated looking thing easy to remember is That whatever is the order of the derivative is the same number for the factorial And it's the same power to which the binomials being raised that work so Might seem like it's a little too much for me to ask today that you would Commit that thing to memory, but you're going to use it enough where it's probably going to be the case anyway And it's not that difficult once you start using it to Generate that each time you want to use it. So this is a Taylor series It doesn't take a whole lot to take what is the Taylor series and make it a Maclaurin series The Taylor series is centered at x equals a we've seen what that means as far as the interval of convergence What it means for it to be centered at x equals a a Maclaurin series is a specific case It ends up looking quite a bit easier It's easier to use because it each time the a value is zero So you have just powers of x as opposed to powers of x minus a so because its uses are I Guess quite a bit different sometimes than just a generic Taylor series It's different enough to have its own name. All right real quick example of how we can Get a Taylor series or a Maclaurin series actually since this is our first one. Let's do a Maclaurin series It's a little easier to do so it's x minus zero or just x to the end So we know ahead of time. We're going to have some higher order derivatives Let's just plan ahead So the original function is e to the x When n is equal to 1 the first derivative. This is kind of easy So all the way down the page All the higher order derivatives are still e to the x that's going to make this Maclaurin series a much easier series because we have the same derivative all the way down the page so for n equals zero it ought to be the Zero derivative, which is the original function at zero over zero factorial x to the zero What is that? What's the first term of the? Taylor expansion. It's going to be one Well, all the derivatives are the same So first derivative at zero would be e to the zero. It's going to be the same In is one so it's one factorial x to the one. What's the value of that term? x n equals to the second derivative at zero over two factorial x to the two x squared over two factorial This is really x to the one over one factorial Guess we're going to get technical. This is x to the zero over zero factorial. What's the next term? If that pattern is going to persist Why is e raised to the zero? Because we're evaluating the derivatives at x equals zero so the exponent of e in each case is zero so e to the x is a Maclaurin series kind of a power series and we don't have enough time today, but we'll Next class see if in fact that's going to generate some of the values that we know it should be generating for e to the x See you tomorrow