 that was so mean he's just so mean well we are back on the air lecture 58 we just did a mean spirited prank on Kelly who I know has an issue about getting here to class and I know why so I don't make a big deal out of it but hopefully you'll accept that in jest let's do one problem that will wrap us up from 8.9 applications to be honest with you what I went what I chose our second example I went through the list of problems starting with 27 on page 629 and the problem that I knew the least about I chose that for an example you don't have to know about the physics or the circuits or whatever it is in the problem as long as we've got enough information handed to us we can take the mathematics manipulate that and hopefully come up with what it is that they want did anybody try the numbers that we came up with yesterday for the error estimate to see if it was 4.17 times 10 to the negative tenth times the mass of the object at rest it not that that's important but if somebody did and validated that that's fine all right let's take a look at problem 27 and when we're finished with this then we'll start to review for the exam by I think since it's the most recent let's take the test that about half of you got back today and others of you got back prior to today kind of work our way through that common mistakes that I saw what might you see on the exam relative to what you're seeing on test for all right 27 let's read it through together and then we'll see some of you I'm sure will feel more comfortable with this content than others try to throw that comfort level out the window if you're handed enough information and you convert it to stuff that is related to what we're doing Taylor series or binomial series we should be able to get the answer in the form that they want it 27 on page 629 an electric dipole can sell when I started kind of glazing over when I got the word electric because I just not very well versed in that electric dipole consists of two electric charges of equal magnitude and opposite sign if the charges are q and negative q and are located at located at a distance lowercase d from each other then the electric field e at the point p in the figure and I'll draw the figure up here is such and such let's go ahead and write that down while we're there so they hand us an equation they tell us what these things are for the most part we haven't gotten all of it yet by expanding this expression for e as a series in powers of lowercase d over capital D show that and here's what we're supposed to end up with show that e is approximately proportional to 1 over d cubed when capital P the point P is far away from the dipole now it doesn't really say in the reading but it does in the diagram that what is capital D it is the distance from this point to the first of the two charges so here's the diagram so here's point P here's the two charges in the dipole the distance between them is lowercase d and the distance of the point which is supposed to be and this is kind of the how we're able to simplify at the end that capital D is a much larger number as compared to lowercase d okay charges are q and negative q located at a distance lowercase d from each other then the electric field e at the point p in the figure is we've got it written down by expanding this expression for e as a series of powers of so we want powers of lowercase d over capital D and that's not handed to us in the expression so we do have to kind of manipulate that a little bit when p is far away from the dipole so when capital D is large compared to lowercase d this is probably this guy right here is probably the expression that we're going to end up manipulating it actually is capital D plus lowercase d to the negative 2 so we're going to use that binomial series which is rooted in Taylor series so we want to convert that so it has a lowercase d over capital D so I think the first expression we can leave alone the second expression inside here let's factor out a capital D so it's inside the parenthetical statement that is in fact being squared so if we factor a d out of here it leaves one that kind of helps the cause because we like the 1 plus x to the k format I'm sorry we are used to using the 1 plus x to the k some of you don't like it and you've told me that which is not very nice at all and when we factor d capital D out of here it leaves one what about when we factor capital D out of here you get the lowercase d over the capital D so it probably is the right thing to do since we want powers of lowercase d over capital D so inside the quantity that's being squared I just factored out a capital D that is also squared right because it was inside the quantity that was being squared so I can't undo that but I can kind of bring it out from this quantity that would probably be the next step so this capital D is squared and we've got a 1 plus lowercase d over capital D squared and by not only factoring out that capital D to get us what we want as far as what's requested of us we've also got a q over d squared right in both of these so it's one it's in there one time and what would be left here one plus over capital D the negative two yes that work we factor out q over d squared this I actually wrote down I guess I didn't write that down that work I'm going to rewrite this because it's kind of sloppy so the directions I know I've read them twice just make sure expanding the expression for e as a series in powers of lowercase d over capital D so I think we're going to be good with that so we're going to use this format by the way if there's any doubt in your mind about whether binomial series will make it to the final exam you can pretty much lock that one in that's a series the expansion of which you're going to need to know for the final exam and it may come in handy on a couple of problems because it is in some problems kind of a shortened version of Taylor expansion if now we had kind of a basic level binomial series problem on the last test on test 4 was 1 plus x to the negative one-third right so if you made an error on that one on that problem look back because there were some errors here's one of the errors is that the factorial was forgotten and then another error is that we've got an x to the one here and we've got larger powers of x as we go and I know that something we hadn't done a great deal of but hopefully you can get that corrected prior to the final exam so we have this thing so let's expand that and then we'll subtract it from one and then we'll see if we can eliminate some of the terms to get this proportionality that's requested at the end of the problem so this ought to be one plus k what is k negative 2 and our x value is going to be replaced by lowercase d over capital D so we're going to have powers of this like we're supposed to have so what do we have here negative 2 and then we've got negative 2 what negative 3 that work and let's get one more negative 2 negative 3 and I I know this happens and this is good that it happens but kind of remind yourself that you've got to generate this from the start I know everybody in here once we've got this term it's easy to pick up that pattern from this point in the problem so you've got to ask yourself can I get this started so that I can easily pick up this pattern that's kind of the crux of binomial series expansion and so on does that look right all right well let's throw this in here let's see if we can take care of all the signs and all the simplification that we need to as we go we're going to have one minus this well this also has a one so the ones are going to drop out right and then everything else that follows is going to be subtracted so this one and this one knock each other out we're going to have what we're going to be subtracting this term we're going to be subtracting this term it's going to be what this is by itself this is positive right but we're going to be subtracting it so negative three and we could leave that d lowercase d over capital D the quantity squared here's three negatives so that one's going to be when it's subtracted it's going to be positive what are we going to have here for do you think that's enough to actually generate the pattern seems to be a two and then a three and then a four if we had to make our guess we would probably say minus five right we're not going to need that checking the directions again when capital P the point P is far away from the dipole so for large values of D we know what that is but compared to the lowercase d it's going to be larger we want it to be approximately proportional to 1 over d cubed if D is large let me put in here compared to lowercase d suppose I truncated this infinite series right there and could I get away with that I guess is the question because if D is really large compared to this one even though this gets squared this is a much larger number squared so these terms lowercase d squared over capital D squared and so on they get smaller and smaller and smaller as you go it may not be quite that exaggerated in the first term but we're going to dispense with those and let's see what we get so I'll go ahead and use approximately because we are truncating the series and what do we get let's see if we get what it is that they asked us to end up with and here we are pretty near the end of the problem and I really don't have any higher level of understanding of these opposite charges and the dipole and the distances and you don't necessarily need it to do the mathematics behind a lot of these problems so I see a D squared times D which I think was part of what they wanted so I see a capital D cubed and a 2 and a q and a D let me rewrite that 2 q D times 1 over D cubed so I contend that we're done with this problem why is that directions are show that e I've got e on this side e is approximately at that proportional to 1 over D cubed well here's 1 over D cubed what if something is proportional to something isn't there a what that interrelates them a constant okay some kind of constant of proportionality so there is the constant that interrelates e and 1 over D cubed capital D cubed so end of problem yes why did you separate the two times D over D from other looking ahead to what they wanted they wanted to D cubed so I already had a D squared so I basically knew I only needed the first term but how can we justify getting rid of these terms D is supposed to be a much larger number capital D than lowercase D so hopefully we're not getting rid of a whole heck of a lot that higher powers of this quotient should be getting closer and closer to zero so kind of knowing where we're headed helps but also dispensing with the rest of the terms we've got to be able to justify that anything else on this problem before we leave it sorry that should be approximately right okay everybody should have their test for back so why don't we begin our exam review with that first one is an alternating series you should prepare for and I don't know right now I've got to sit down and write this exam out based on what your four tests are and what we've covered since the final exam since the final exam you can pretty much lock in a binomial series problem a little more complicated than the one we had here because this is just kind of bare bones 1 plus x to the k so you may actually have to manipulate it a little bit first to get it in that form make a substitution for x and then make a decision what your k value is not that much worse than this but be prepared for one just a little bit more involved than this probably not any of these applications because most of them are kind of lengthy what we did yesterday and the error estimate took pretty much the whole class right I guess I could just have one of those on the exam and that'd be the whole exam that's not a good idea so probably none of those just because of the time consideration but I think they having been through them might have a net effect of making the binomial series that we do have on the test a little bit more mundane for you because we've done some that are worse now back to question one alternating series you should be able to recognize that as an alternating series whether it's in its expanded form that's easy they alternate in sign or in the closed form the sigma notation form that negative one to the end is a pretty dead giveaway and then you have to do pay attention and I think this got a couple of you I didn't intend for it to trip you up but the first series starts at one n equals one and not n equals zero so I think that got a couple of you and they lost a couple points because you started yours at zero which generated an extra term and made the sum quite different so pay attention to where it starts negative one to the end means it's alternating we do have a special convergence test for alternating series two parts how does that go okay the n plus first term is smaller than the predecessor the nth term and again we're not we're throwing out the negative and positive alternation stuff so kind of the argument the rest of the argument other than the negative one to the end so is that true it means it's ultimately decreasing now you've what you can't do and I think I had one person do this maybe two people on the test is you can't just write this down and say you know check yes you've got to make it specific to the problem that we're doing so in this problem you've got the solutions if you didn't get it correct most of you wrote that step out and I think it's some of you even wrote out why it's true because you have a larger denominator which makes for a smaller fraction but if you wrote it out and you've justified it and you've moved on to the second point then you're good but you got to make it specific to that problem and the second one other than the negative one to the end so the the other part of the argument if we work our way is that zero so are the terms disappearing as we work our way out to the right not quite as stringent really as some of the other tests but because it's alternating these two things if it's ultimately decreasing in the value of the nth term goes to zero that's good enough we've seen series that the alternating version is convergent and the positive term series that is its counterpart is divergent namely what the harmonic series the regular positive term harmonic series is divergent its alternating counterpart is convergent so it is not quite as stringent of a test because of the fact that it's alternating take the series write a few terms out if we are finding the sum if we want to deal with a certain error tolerance for the sum where do you search for the first term to delete if you've got a certain error tolerance where do you want to stop it and you want to add the first four add the first seven add the first eleven how do you make that decision okay wherever you're going to stop let's say you're going to stop at n equals four then you want to look at the next term right basically the first omitted term if it's smaller in magnitude then your desired level of accuracy ignore the sign then that's good you can delete that one gather up the predecessors of that one add them up and there's your desired level of accuracy so that should all be shown on the solution to the test that was handed out so you don't how about you needed to use that one yeah the example we did in class we used the term that had like the four zeros and five zeros because like your answer isn't correct in four decimal places because the next one will take away one of the decimal places in that one okay so we ask what this is good this because it five made a mistake in grading today's a great day to catch it so what was the level of accuracy that was desired four decimal places so i guess we have to ask ourselves the question what what decimal place could potentially affect the fourth decimal place so i flag this as the fifth term which is point zero zero zero three four three is there a chance that adding that or subtracting that could affect the occupant of the fourth decimal place yeah actually that there is right yeah because if the occupant of the fourth decimal place let's say was two point zero zero zero i'm sorry the occupant of the fifth decimal place was two point zero zero zero zero two and we added this in that would make the two a five and then we're talking about it affecting that right so that that is correct so please change that that we should actually gather in the fifth term here because there is a chance that that number even as small as it is and it has no nothing in the fourth decimal place adding it or subtracting it could affect what occupies the fourth decimal place so if that causes me to add back in some points see me at the end of class and i'll write that down so it gets changed uh but we should actually go on to the sixth term which should become the error term and that should make the sum i guess it could make it potentially two eight what is this term subtracted fifth term is subtracted negative point two eight three five okay i think that's correct so if i need to change that see me when we're done today and i'll change that so that question that we need to ask ourselves is could that decimal affect the accuracy at the desired level and in fact this one could thank you for bringing that up okay number two um it is alternating by the way um well let me scrap that this one because of the x powers of x plus two probably is an indicator that and plus we're searching for the interval of convergence and the radius of convergence ratio test is written all over this even though it's not actually said to use that certainly you should plan on using the ratio test in some way shape or form on the final exam we used it a lot in chapter eight um and there's often there's a lot of stuff to keep track of in this problem you've got powers of two in the numerator and the denominator you've got powers of x plus two in the numerator and denominator and you've got an n in the numerator and an n plus one in the denominator so if you slash something and mark it out when you're looking at x plus two by the way all this is in absolute value right so when you start getting rid of things uh for example the twos remember that there's going to be a two if one of the powers of two is one larger than the other one there's going to be a two somewhere where's the extra two going to be in the denominator so either if you do mark out the exponent and you want that two to remain you better circle it or something so it doesn't get lost because there's a lot of stuff that's going to be slashed here and reduced or simplified we've got powers of x plus two here powers of x plus two here where's the x plus two going to remain we've got one more of those up here so you've got an extra x plus two and the other one is n approaching infinity what happens to n over n plus one that approaches one right so a lot of stuff going on so make it clear what's remaining and where it's remaining so you've got x plus two over two and if you want to call that absolute value of x plus two over two that's fine but that's the ratio and we want that to be less than one if this is supposed to converge and some of you i know could kick yourself because you did all the hard part right and you multiplied by one half at this stage multiplied by one half you've done all the hard work now we've got to do a little arithmetic don't think multiplying by one half is going to help the term that's in the middle don't we want to multiply by two right a couple of you had your inequality symbols kind of out of whack there too you might want to check those all right so ratio test interval of convergence we would add to and i'm sorry subtract two so our initial kind of first look is negative four to zero and a couple of you stopped there you're not quite finished on a ratio test problem and determining the interval of convergence because if x were negative four back up here to the ratio test we'd get one so the test fails if x were zero back up here to the ratio test we would get one and again the ratio test fails when the limit is one so we have to do those by hand so check those out i think it diverged right at negative four and it converged so our final interval of convergence looks like that and the radius it is centered at negative two which that is that interval and the radius of convergence is two uh third problem power series we also did a lot with power series there's a whole lot to choose from from this test and we even though we weren't progressing through the text real rapidly the latter part of chapter eight there was a lot that was in there and all the power series stuff we used over and over and over again in fact the first time we saw that was in this form right and we saw it in this form initially what is that where's that from infinite geometric series now we can't say that it converges to this until we know what else is true about that infinite geometric series so if it's negative three fourths we're good if it's positive two thirds we're good and if it's not something we can actually quantify in terms of a numerical value then we have to then find the interval of convergence so in this problem most of you did the problem this way where you saw the nine and you did not want a nine there you wanted a one there so you divided the numerator by nine and the denominator by nine a couple of you just pulled a one ninth out in front and then you reinserted that at the end of the problem that's fine so if we want to put it in this form we get the one in that position and then we if we don't have subtraction we convert it so that it is subtraction so that ought to be the first term and that ought to be the ratio so if you want to write it out and about half of you wrote it out in expanded form and half of you wrote it in the closed sigma notation form it's fine doesn't matter as long as they're equivalent so the first term that's a giveaway that's just whatever is in the numerator and if you pulled a one ninth out your first term would be x but the one ninth times that x would eventually be x over nine so the end result is the same so we would then take that first term times the ratio first term times the ratio again so the ratio will have been used as a factor twice by the time we get to the third term it's always one behind so if you wrote that out hopefully simplified it it looks like the denominator there's nine there's a nine squared so it looks like nine nine squared nine cubed looks like the denominator is just powers of nine and in the numerator what's it look like as you so the negative squared is positive x squared squared is to the fourth times this x is x to the fifth and this is nine times 81 729 is that right so if you wrote that fine I think I've said that that on the directions doesn't matter expanded form or closed form but if you do write it in closed form probably be a good choice on your part to write it correctly in closed form uh some of you kind of started from this first term times the ratio to the nth as long as you started that at zero we can get away with that because the ratio doesn't really come into play until the second term of the series and if you started it this way there's an x um I would recommend taking the negative one to the n out of here which leaves what x to the two n is that right and we've got a nine to the n and then we've got another nine down here does that take care of everything that's present to the right of the sigma notation uh x to the what two n plus one by the way two n plus one if you double a number and then you add one to it it's guaranteed to be what odd right so two in by itself is even two n plus one is always odd and in fact that's what we have in our series and what nine to the n plus one if we start in at zero I know some of you started in at one you still got it correct just kind of adjust back from what we have here I don't personally I don't see a difference in this answer in this answer it's just kind of your preference of how you want to write it uh we talked about number four about the hyperbolic cosine the cosh function uh we also threw in the hyperbolic sine they are very closely related to the regular circular trig counterparts but they have to do with points located on a unit hyperbola as opposed to a unit circle uh five on this test was really not meant to be a difficult question I really kind of thought it was almost like a gift question but many of you rejected my gift um so I think we talked about going into the test that it would be good for you to know and I I still stand by this statement I think it would be good for you to know this you could develop it but it's going to I think be a waste of your time sine of x we decided sine was an odd function right so we should have odd powers and odd factorials x is really x to the one over one factorial it is alternating and I know that we don't want all of mathematics to be this memorized tremendously large body of memorized facts but it you know some of it comes in handy if you're going to memorize some of it it's going to save you some work uh that's the sine of x in these x's that appear on the left side of the equation and the right side of the equation are the very same thing so if we want to take that and now talk about the sine of instead of x the sine of x squared everywhere there's an x there should now be an x squared and we've talked about that we did some problems like that and the real goal the problem is to integrate this now we have a power series and I didn't really address this in the problem do we have a choice do we have another way of integrating that what about if you let u equal x squared du is 2x dx aren't we out of luck with that method because we don't have an x anywhere in the integrand so regular substitution in fact I don't really know that we have another method for this so if we want to integrate this side which we do we can integrate this side and accomplish that excuse me so it should be the integral of that x cubed over three this is really x to the sixth right so if we want to integrate that it's going to be the coefficient that's already there one over three factorial x to the seventh over seven here's an x to the tenth integral of x to the tenth is x to the eleventh over eleven and so on so we know or develop fairly quickly the sign of x it's Taylor series we then replace x with x squared or something else I mean think about what I could do for an exam question here it instead of x going to x squared it could be x going to x over five or 2x or negative 3x and then we could integrate that new function by integrating each piece of its expanded power series now we are doing an indefinite integral and every time we do an indefinite integral we should put a c or k value and I don't think we have any additional information so that we're going to be able to find that c or k value uh real quickly number six is very similar to this again in your library of functions I think it would be and this one by the way if you don't remember this this takes I'm sure less than a minute to develop because all the higher derivatives of e to the x are e to the x you want the higher derivatives evaluated at zero e to the zero is one each time so it's going to be basically just the other part of the Taylor expansion is all you need generically this would be handy this is I guess more specifically a McLaurin series but still is Taylor because we're evaluating at a certain value just the value simpler so a couple of you um on the fourth problem made a mistake here and hopefully you'll get that corrected prior to the final exam you did find the higher order derivatives but you in you evaluate each of the higher order derivatives at zero not at the end value that we are happened to be on at that particular point in time in the expansion so if you need to use that to develop it it doesn't take long to develop this one that's e to the x how about e to the x square what we just do in going from sign of x to sign of the quantity x squared just replace it right on the expanded version and then we want to multiply that by x squared just run an x squared through there and again some of you did this in the closed form and may have actually gotten there a little bit quicker we've got an x squared times everything let's figure out what e to the x is well normally e to the x is x to the n well now we've got x squared to the n over n factorial and what do we do with that x squared that's out in front we want to pull it in here doesn't that add two to each power so this is x to the two in I guess we want x to the two in plus two does that work well the first term is x squared when the n is equal to zero is that what we get the next term is x to the fourth when n is equal to one is that what we get tell me something about two n plus two and then we'll stop today even and how are the exponents going to progress they're all going to be even and they're going to go up by two and they're going to go up by two because you're doubling in before you add something to it so anytime you double it you go up by increments of two if you triple something you go up by increments of three and so on and again I don't care if you use this one or this one those are equivalent to me all right we won't meet tomorrow we'll be back in here on Friday bring the rest of your tests if I had a correction to make on your test bring that up and we'll take care of that now