 Let me remind you, this stuff you already know, final is a week from today, late at night. Most of you should go to the Union Auditorium. There's a problem session Friday afternoon at 2.15 over in Jadis. And today is a Wednesday, even though it says Monday on your calendar. So we still have a lot of stuff left to review. I'm not planning to review differential equations unless people want me to because I just did that. So really what will be left off in the review was volume. So I don't need to cover volume anymore, right? I don't know. Series? So let me just remind you what we're skipping. Work. Work is almost the same as volumes, except for the setting up part. So it's the integral of the force over the distance that the force is applied. So most of these problems that people start, finish. So a simple work problem would be something, so a very simple version would be something like, you have a force which is, I don't know, 3x squared. And then you want to move, so x is set in distance. You have some, let's make it 3 over x squared. So that would be something like we have some object here. It's a magnetic charge that's repelling. We have a magnet that we want to move in here where x is the distance. And we want to know how much work it would take to move it from 3 units away to 1 unit away. So this would be very simple. It would just be the integral from 3 to 1. The reason it's negative is because the integral will be negative, but again it needs to be positive of 3 over x squared dx. So this would just be, what is that, 3x to the minus 2. So that's negative 3 over x evaluated from 3 to 1, which is negative 3 minus, yeah, minus 1 plus 1. Okay, so something's wrong here. Well, yeah, except the force is the other side. So then that gives me, no, that was right before, wasn't it? Yeah, right, 3 to 1. From 3 to 1, so this is negative 3, you can see the same thing. Negative 1 plus 3 is 2. So it takes 2 whatever the right units are. So these are simple. The idea is exactly the same. The difficulty that people always have is calculating the force. So if I tell you the force and set up the problem, you just integrate. That's easy. The difficulty that people have is what is the work that it requires to pump water out of a tank? Or what is it the work that requires to stretch a spring or something like that? So that's where the hard problems come in. But the idea of work is very simple. You're just calculating how much effort is this, how much force do you need to do that and so on. So instead of a problem like this, you tend to see things more like, I don't know, would you prefer moving something or a spring problem or what? So what do we do? So we do something from this. Good thing. Alright. Do you want a water pump? Okay, so how much is needed to pump, I don't know, water or other shaped tank, which is, let's say, height equals, I don't know, 12 meters. The radius, let's say the diameter at the top is, I don't know, 6. And let's just say it's completely full. Well, let's make it a half full. So it's full to a depth of 6 meters and we need to know that water has a density of 1 kilogram cubic meter and we need to know the force of gravity is 9.8. Do we need to know anything else? That's really all we need to know. So this is really a jazzed up volume problem where we're computing not exactly the volume of this surface of revolution but we're computing the force that we have to use to lift something. So in terms of the picture, this distance is 12, we have 6, here we have 6. We have this much stuff and we want it to go out. You want the spout to extend or you just want a spout where we pour off the top? Extend. How long would you like the spout to be? 2 meters. Okay, I don't understand why that makes it harder but it seems to. Okay, so we have a 2 meter spout. You don't understand why the spout is difficult but it seems to. Okay, so what do we have to think about? Well I'm just going to draw the same picture again but in cross section. So this is 12, I'm going to now label this as 3 because twice 3 is 6. If we have a piece, a little slab of water at height y, what we need to calculate is how much some water at height y needs to travel. Here's my spout, that's 2. So how far a slab of water need to go? That will tell us everything because we need to know the force and we need to lift this so how heavy is it and how far do we have to move it. So this little slab of water at height y looks like a little disc. Of course the sides are sloped but we can ignore that. A thickness dx or dy. The radius here, we don't know the radius we want to find that, that's r. And we can use the fact that it's at height y. So we have similar triangles here. The height of the tank, since it's a cone, the height of the tank to the radius of the tank is 4 to 3. So that means that this little triangle here is of height y and, why did I say 4 to 3? 4 to 1. So the radius is y over 4. So we have r is to, how about y is to r is the same as 12 is to 3. Because these are similar triangles. The distance from here to here is 12. The distance from here to here is 3. The distance from here to here is y. The distance from here to here is r. So the ratio of this to that is the same as the ratio of this to that. So that means that cross multiplying r equals y over 4. If we had a different shape, we might use a different method to figure it out, but okay. And so that means that the volume of this is dy times pi r squared. Volume of this little thing. And the distance it has to go is the distance from the very top, which is 14, from y to 14. Play us a little here. This is really 12 plus 2 more for this bout. And so that means that our integral, well, we integrate the mass of the stuff, which, since the density is 1, times pi over 4 squared, times the force through the gravity, which I guess is 9.8. So this is the volume. This is the density. Huh? I'm not doing any grams. Kilograms. Kilograms. What? Okay. Then it's 1,000. 9.8. Do you want? Oh, no. The distance it has to go, which is 14 minus y, dy. And how much do we have to move? Well, we have to move the stuff at height 0, all the way up to the stuff at height 6. If it were a sphere, we would have a different formula, which would mean that we would calculate this little slab r differently. If it were a pyramidal box, then we would have a different formula for the variation of the thing. This is exactly the same concept, exactly the same concept of finding volumes of surfaces with a known cross-section. The only difference is we have to multiply by the distance it needs to go. That's it. It's exactly the same. Stuff with the spring, exactly the same. You just integrate the force into the distance it needs to be applied. Okay? Do I need to say any more about work? Yeah. How come we didn't split it into two integrals from, say, 0 or 12? Okay. If you'd like, I can integrate from 0 to 6, and from 6 to 12 of 0. I meant, and then from 12 to 14. Plus the integral from 12 to 14 of 0. If you want. There's nothing you have to do. There's no water above the height of 6, so it has nothing to do. So let's make the problem harder. Suppose this is a mixture of oil and vinegar. So the oil is at 6, and the vinegar is from 6 to 12, and we have two integrals. We have one for the work to lift the oil. Oh, oil flow. Sorry, the vinegar is at the bottom? No. Yeah. The vinegar is at the bottom, so we would have a different density. So we would integrate from 0 to 6 at the density of vinegar, and then the density from 6 to 12 for the oil. But other than that, it's the same. Yeah. Because there's no water that needs to go out between here and here. So look at, just consider one little slab of water. Okay? So it's hard to think of a slab of water, so think of this as plastic. I'm going to shave it up. So I chop it up into a bunch of little slabs. And the work to lift this slab from here to there is this. The amount of work, the amount of effort I need to expand to lift this little cylinder at height Y is it needs to go from here to here. That's 14 minus Y meters. Yes? Okay. From here to here, 14 minus Y meters. How heavy is it? Well, it depends on how big it is. How heavy is it? Well, it's this times this times that and then I need the 9.8 from gravity. So that's how much work I need to do just for this skinny little slab. The integral is letting me out of all the skinny little slabs. So there's a slab here at height 1. There's a slab here at height 1.5. There's a slab here at height 5. Each of those causes me to do a little bit of work. Up here there's no slab. There's no work to do it. That's where this comes in. If I put the spout right here then there would be no work to take out this one. And in fact, if I put the spout here then there's no work at all. Just open the tank. It's negative work because I can put a little generator on here and get some energy out of it. So integrals let you add up things. What am I adding up? I'm adding up the energy, the work I need to do a small thing. I have a lot of small things. I need these slices. So I'm adding up the work for each of the small slices. The integral is not telling me the amount of work for any one of the slices. The integral is letting me sum up the work for all of the slices. So I still continue as I can see. Okay. Do I need to say more about this? I don't think that will help. Okay, let's move on. We talked about our length. We talked about average value. I think we've done all of the all of that stuff. So we can move on to zeroes. Suppose I should say something about, do I need to say anything about sequences? They are just limits. So we have two objects. One of which we pay a lot of attention to. One of which we pay little attention to. So if we see something like this this is just a list of numbers. So something like two equals three to infinity just means one-third, one-fourth, one-fifth, one-sixth, just a sequence of numbers. So this is called a sequence. And the typical question is does this converge? So that's really just a limit question. So if we say what is the limit of the sequence we're in? Then you calculate the limit as n goes to infinity of one over n which is zero. But this guy converges. The only difference here is these functions only take on integer values. So if we asked about something like the limit of sine of 2 pi n well that's zero even though the limit of sine n doesn't exist. This is easy. Anybody have any? So the biggest confusion that people tend to have is they answer the wrong question when they're asked about a sequence. They say, oh we have to use the ratio test or something. Those are for series. So an object we have is sequences. This is a tool we use to understand infinite series or sums. So this might make an appearance. It's small even though it's fundamentally important. So more something we spend more time with is the idea of infinite sums which are also called infinite series. Series is the more common word. I prefer the word sum because of the meaning in English. Sum makes it clear you're adding. Series is confusing to people but too bad this is the word that people use. And we tend to write them with sigma something like one third to the n. Something like that. So for infinite sums so this means take n equals zero so we get one third to the zero which is one. Then take n equals one which means we get one third. Then take n equals two which means we get one ninth and take n equals three we get one twenty-seven and keep doing that forever. And then the question is does this add up to something? That's a nine. Does this add up to something or does it get big or does it do weird stuff? That's what we're trying to solve. So for series this one that I wrote down I don't know how to add up so for example the geometric series looks like so one that we have the formula for starts at zero n equals zero to infinity some constant is it usually a? I guess it is. Some constant times the ratio to the n of power it's not hard to figure out the formula that this is a over one minus r if r is less than one in absolute value so this one a is one and r is the third so this guy adds up to one over two thirds which is one and a half. You definitely should know how to sum the geometric series. It's important to remember the geometric series started at zero so if you see a geometric series it doesn't start at zero and you either need to adjust a to absorb it or somehow otherwise manipulate it to make it look like that. There are also sometimes telescoping series which would be something like so let me remind you really what we're doing with an infinite sum is we are calculating what do we get when we add the first term what do we get when we add the first two terms what do we get when we add the first three third terms so we turn this series into a sequence a sequence of partial sums so in general what we're looking at the sequence sn which is I add from let's call this sk which means I add starting at zero or whatever the first term is a to k so I take a0, a1 a2 and I stop at the kth term I call that sk and I want to say does the resulting partial sum does the resulting sequence does this converge that's really the question we're asking when we're looking at an infinite sum we want to add things together converge so for a telescoping guide we might have something that looks like let me leave it factored looks like this each term kills off part of the previous term if we write out terms for this we get one half minus one backwards so this one doesn't work this one's not telescoping so then I get one third minus one half maybe it does then I get one fourth minus one third this is okay and so on and what happens well if I stop at any stage notice that well I have this minus one always half so if I list the partial sums s one is negative a half s two is well if I just take the first two then this half kills that half leaving me minus one plus a third s three is minus one plus well this that third would be the fourth and its place and so on and if we look at the last s k well not the last but if we go for a long time we get minus one and the half kills off and the third kills off and the quarter kills off and then we're left with the one that didn't get killed off so if we look at the limit as k goes to infinity of the partial sum this is the limit as k goes to infinity of minus one plus one over k which is just minus one so we have just the first term yeah yes you're right doesn't make a difference the limit is still the same I just wrote it down wrong right minus one plus a quarter s k is minus one plus one over k plus one so thank you for fixing that but silver limit is minus one of course I can fool you in some sense by making something here where this last term doesn't go to zero right where it still kills off the earlier terms but this thing is always non-zero silver limit only exists because maybe this thing will grow so you have to look a little more carefully you can't just say oh look stuff cancels out so for example this guy converges to minus one but if we look at well I don't have one in mind right now alright so we have these telescoping guys that may cancel out but in general it's often hard to find some problems directly so we develop just like we did in for integrals and just like we did for derivatives we develop techniques to put things together so the various tests we do to see whether things converge might be the integral test so if we can't add it up we might just ask does it add up? and then if it does add up then we can approximate it by some so for example one thing we develop is the integral test this says that if well let's just say instead of if let's just say the sum from n equals whatever of an from whatever an so just you look at your term and you integrate and these do the same thing if the improper integral which you get when you integrate the nth term convergence then so does the sum if the improper integral divergence then so does the sum the integral test conceptually it's pretty easy just integrate the term I guess we also see let's call it a divergence test I should have said first if the limit as n goes to infinity of an is not zero then the sum diverges if it is zero you don't know but if the limit of the nth term is not zero then the thing diverges if I'm ultimately adding one all the time then I'll get a lot of stuff the integral test here this is for positive terms non-native terms I wrote integer I wrote it on the board I can't list also good so do I need to probably leave that up for a minute one time I'm sorry you need me to list all these tests yes keep going well because we have fun at a time but okay I'll keep going so we have the integral test for once another one that we have is say the comparison test so again this is if we have a series of positive terms then if we have if we have something smaller that we know about and that smaller thing diverges then so does the bigger thing if we have a bigger thing bigger thing converges so does the small thing now we need to have some things in hand to compare to so one thing that you get right away out of the integral test is a class of series called the p-series one over what am I doing we write it as x to the p or one over x to the p I guess it doesn't matter so here if p is bigger than 1 if I write it this way from n equals anything to infinity what did I just do wrong here one over n to the p that's the problem here if p is big then this converges if p is small it diverges if we have a sum of powers if we have a sum of powers and those powers are big benefit converges that follows immediately from doing the integral test on this you see that if p is big you get a number if not you don't and so that gives us a big class of things to compare with for the comparison test unfortunately the comparison test doesn't always is it always easy because it's sometimes hard to locate something to compare with so I should just do two quick examples so for example if I want to know what about the sum n equals 2 to infinity one over n q plus let's see if I want to plus here so that's smaller so if I wanted to know about this does this converge or diverge if I use comparison and what do I compare to one over oh yeah it's a cube so we compare with one over n cubed does that tell us whether it converges or diverges so this converges two reasons since one over n cubed plus five is always less than one over n cubed and sum of one over n cubed is a p series and if I have something like to change that to plus two minus I can't now use that but it converges so I can compare to one over n squared but I can always say well this looks like one over n cubed a whole lot so there's a version of the comparison test called the limit comparison test which says that if I look at the limit of the ratio and this is not zero or infinity and a and d n do the same thing if you know if the ratio of the things looks the same for big n then they do the same thing so here in this case I would use limit comparison at the limit as n goes to infinity of one over n cubed over one over n cubed minus five that limits one and so limit comparison says that the sum over one over n cubed minus five converges just like one over n cubed because that's a p series it converges I guess the other test that we had is so all of these are for positive term series sometimes the terms are not positive so we might have an alternating series this is one that designs alternating and alternating series are particularly easy because you can use the alternating series test so the alternating series test says if the absolute if the absolute value of the terms so the series is alternating plus, minus, plus, minus plus, minus and the absolute value of the terms well that are decreasing this way terms to zero then this converges one other thing we get out of the alternating series which we also get out of the integral test but I can write it down is not only does it converge but we know if we stop after k terms where we're off so for the alternating series we get that if we add a alternating series only and stop somewhere if we went up to infinity by no more than the term we didn't use for example we put the both together let's do the most straightforward alternating series the alternating harmonic series so this is minus one plus a half minus a third plus a quarter minus blah blah blah this guy converges even though if I change all of these two pluses it diverges this guy converges by the alternating series test since one over n plus one is less than one over n and the limit is n goes to infinity one over n is zero so it converges by the alternating series test and if I stop after 100 terms so I do minus one plus a half minus a third plus a quarter and I do it for a long time and I stop at one over 100 then this is off from if I go to infinity not over 100 anymore because the term I didn't use controls everything all the rest is smaller than that all the sum of all the rest okay so this was a setup to be able to do Taylor's series and McLaurin series which I have to cover now in six minutes Taylor's series I guess oh I missed the ratio test I'm sorry don't forget the ratio test let me write it here if the limit is n goes to infinity of the sum not the sum of the next term over the current term is less than one so if the ratio of the next term to the current term is less than one then this thing goes faster than the Taylor's series so the series converges and in fact it converges absolutely which means that both the series and if we change some of the sum it still converges if this limit is bigger than one then the series diverges and if it equals one we have to try something else so the ratio test gives you no information when the ratio is zero when the limit is zero I mean is one you have to limit comparison which was somewhere is here which works when the ratio is one so you have to keep track which is which but okay so the ratio for me the two most useful tests well alternating series because it's easy so the three most useful tests I find the ratio test limit comparison most useful unless there's something obvious like an alternating series or something like that if I'm just handed a test a series out of the air the first thing I try is I say can I compare it to something let's try limit comparison if I don't see what to compare it to I try the ratio test if that doesn't work okay so let's see if I can get Taylor series now in one minute so for power series we can or for author we can write a function as a sum of something involving x let me just write it this way for a minute so we can write some function where the terms all depend on x usually this will be a power series that looks like some number times some power of n so we can do that the nice thing about say Taylor series or McLauren series tells us how to find these coefficients so McLauren is just Taylor series it shouldn't have used c oh well it's expanding around zero it tells us that the function is well it's its value at your favorite point let's call it a plus the derivative at that point times how far off you are from that point plus the second derivative at that point divided by two how far off you are squared plus the third derivative at that point I guess I go on to the next board so plus the third derivative divided by three factorial by how far off you are cubed so in general the nth derivative over n factorial by how far off you are to the nth power and then we keep going so from McLauren series a is zero so you can do this for a lot of series you should know definitely should know McLauren series for e to the x for the sine for the cosine for the geometric series say for the log this one you can get from that this one you can get easily from that one that's pretty much it so if you know these you can get pretty much all the other ones we can get them by say integrating and differentiate could be on either depending on how I say it so for example a reasonable thing on the easy part would be tell me the McLauren series for sine of two x because you know you know the series for the sine of two I mean for the sine of x and you put a two everywhere you see an x something harder might be give me the Taylor series for the sine of two x expanded around x equals pi over four that's part of it or of x squared sine two x x over x or something where you have to do more work so the easy part would be tell me one of these or a minor modification of it the hard part would be a major modification amount of time so what I recommend you do since we're out of time is if you have more questions I will be around but I'm not scheduling office hours this week if you want to come see me give me a call set up the time post questions on kiazza stuff like that I'm holding the review session problem session on Friday I'm done reviewing now if you want more reviewing you can go bother with this one