 This is the distribution on the exam. People, there weren't a lot of A's or A minuses, but in general, people did sort of a little better. There were a lot of B's of various flavors this time and last time around. There were also, unfortunately, a lot of people in the set face category. So the average is just under 74, the median is really a little more useful. He is 76, so the median means as many people got above 76 as got below 76. So the mean usually tends to be a little lower because of these kind of people who bring things down. So actually that one is a zero, but I don't record a zero because it's a little hard to tell whether a zero is just a forgotten score or a zero. So I gave this guy one point. Okay, so that's how things went. Let me clarify a couple of issues regarding grading in this class because I think there is some confusion. It's a little bit complicated to explain exactly how I do the averaging. It is a straightforward way, but every time I try to explain it to students, their eyes go funny and they don't understand what I'm talking about. I take each of the individual scores. I put them on a standardized scale. I'll tell you what my standardized scale is because that's one of the things that increases people, but I put them on a standard scale and then I compute the weighted average, which means that if you're talking about midterm, that counts sort of 25% total. If you're talking about the final, that counts towards 35% total and so on. And that gives you a number. And that number is your grade. It's not exactly your grade. It is the minimal possible grade that you can get because in some cases, so for example, if somebody really figures things out so they get an F on the first midterm and then things start to come together for them and they start getting Bs, well then I don't want to give that student a low grade. So if there is significant improvement, I take it into account. That does not mean that your grade is the best of your average or your grade on the final. Should be? Okay, however, one thing I will not give a grade of lower than C to anyone who earns a decent C on the final or better. So even if you have solid zero going into the final and you figure things out and you earn a C or better on the final, I will give you at least a C in the class. If you were right on the line between C and C minus, that doesn't count. So if you just barely clear a C on the final by one point, you're back. But the line of C on the final means demonstrated mastery of the material. So if you demonstrate at least passable mastery of the material on the final I will not give you a useless grade in this class either. So you need at least a C to go on to the next class or to have this counselor with your major or any of that sort of thing. Well, that's more amazing stuff. You need at least a C. So if you earn an honest C on the final, I'll give you a C in the class. It does not mean that if you earn an A on the final, I'll give you an A. I will look closely at your grades and make this a C. Okay? Yeah. Any other questions? So who at least needs midterm to back the grades that you have already? You can check them on the web. You can get them back again in presentation. And please look at your exam for any addition errors or grade errors or that sort of thing. The solutions are up there. Yeah. You have a question? Okay. Well, then that means that you've gotten this report. It's part of why I report the grades. Because mistakes happen. So if it says your first midterm was not there and you got an F, or if it says your first midterm grade was a 27, yet you have a piece of paper that says 105, this means somebody typed in the wrong number when they reported your score. So bring your midterm, your paper to me, or to your TA and say, look, this says I didn't do it, but here it is. The same thing if we added 10 plus 10 and came up with 10, please let us know. These things happen when you're adding up 350 papers at 3 in the morning. Then this is a little more problematic. So if you can't find your midterm, you have a record that you took it because you signed your name. So it's a lot easier if you can find your midterm because then we can say, oh, this missing grade, it was 175. But if you can't find it, we can't know what it is. So that means we have to do something else. If the grade isn't reported, it's our fault. It'll do something really better before it's corrected. Any other issues, questions about grading a jump like that? Okay, so let's go back to doing math. This idea that if we have a function, we can determine the power series for it by evaluating, as an example, we did the series for e to the x and near x equals 0. So if, by looking at the derivative, we can determine this is, which is x, so that e to the x, well, the power series of e to the x that we worked out is the powers of x divided by n factorial. So we went through the, I'll go through this again more, but we did this just sort of by hand to determine that this power series is converted to this function. And we also, at the end of the class, the sign of x is x plus x cubed over 3 plus x to the 5 over 5 factorial minus x to the 7 over 7 factorial plus blah, blah, blah, which is n equals 0. It's an alternating series which alternates every other time, so we only get the odd powers of x and then the odd times. Now we can go through the same process to determine a series for the cosine, but we can also just determine a series for the cosine by taking the derivative. So if we want, given these three facts, we can go through from first principles and figure out that the series for the cosine should be 1 minus x squared over 2 factorial plus x to the 4 over 4 factorial, et cetera. But we can also just take the derivative which gives us the cosine and just take the derivative of each of these terms and see that this is 1 and then this 3 comes down, giving us x squared and the three principles with the 3 factorial leaving us, I guess, 2 factorial is 2 and then for the 5 term the 5 comes down and gives us a 4 factorial so that gives us x to the 4 4 factorial x to the 6 over 6 factorial and so on. Or 4, the microphones deal with. So, or if we look at this we'll see that this gives us x to the 2n over 2n factorial minus 1 to the n so the first term falls away and then we re-enter this. So you can do this just by taking derivatives or you can do this just by starting from knowing nothing and seeing what happens. So let me encapsulate, so all of these things these are called, it might be an i I think it's an i. McLauren's series named after the Scottish mathematician who figured them out are also there called Taylor series. Taylor series is a more general thing more general version of McLauren McLauren series came first but Taylor series is more general I'll talk about what Taylor series is in a minute and let me just remind you the process but in general So to figure out the Taylor series or the McLauren series so the McLauren series is the Taylor series I don't need this So let's say Taylor series So the general process here just to codify the business that we went through last Friday I have a function f and I know f of 0 f prime of 0 f double prime of 0 all the derivatives at 0 and then what I can do is I can determine that f of x is the same thing as first evaluate the function at 0 that's the first term of the series then the next term of the series will be what I get when I evaluate the derivative at 0 and I multiply by x and then the next term of the series will be what I get when I evaluate the derivative at 0 but I have to divide by 2 because if you remember this process we went through to determine this the way to determine this is we say we have some power series 1 of x is 0 then f of x equals f of 0 so that's good and then we take the derivative to kill off the constant term and turn the derivative term into a constant term at 0 but here when I take the second derivative I get a second I get a power of 2 so I have to divide by 2 and then when I take the third derivative I have to divide by the 2 from before but then also the 3 take the fourth derivative I have to divide by the 4, the 3, the 2 and I just keep going like that so in general I take the sum from 0 to infinity of the nth derivative of x evaluated by 0 divided by n factorial this gives me the coefficients this is a number and then I multiply by x to the n so this is the form for McLauren series I just keep taking derivatives and plugging in that gives me the coefficients and this gives me the variable so this is only true of course if this series converges so we still have to check the radius of convergence and that sort of thing that's why we went through this business with the radius of convergence when this makes sense before we look different the way to understand the function essentially what it does is it lets us trade every detailed local information we know what happens at 0 we know derivatives at 0 we know second derivatives we know all the derivatives at 0 and knowing what happens at one point really well tells us what happens nearby and we can go the other way knowing what happens nearby tells us all the derivative information at one point so we can either trade knowing sort of high level information in an area for knowing very detailed information at one point this is if you think about it not really new to you but it kind of is you haven't seen this I mean you probably haven't seen this before stated in terms of infinite series but you actually did see this in first semester calculus a little bit so in first semester calculus one thing that you did a lot is that I had some function and I said well this function is just about it's value at some point plus the derivative times I mean usually you see this with an a here but you've seen this before this is the tangent approximation this is just saying if I know here's my function not much of a function but if I know the value at a and I know the derivative at a I know a line that approximates it so this is the equation of this line and this is the graph of the function we've done it is there a question? and then all we're doing is we're saying well we don't want to align here why don't we use a parabola so if we use a parabola and we want to find the parabola that goes through here then we would actually have a parabola that goes through and so in fact let me use a little technology here but only in this room does it not work alright so too bad so what I'm going to show you was the convergence of this theory to the sign so I will have to use this kind of technology so there's the graph of the sign that looks like the graph of the sign here's the graph of the sign and the picture that I was going to show you was if we take just the first term we get that if we take the second term of the series we get something that looks kind of like this if we take the third term of the series we get something that goes kind of like this and as we take more and more terms we are looking like the sign think about polynomials of higher and higher degrees we don't get more and more bumps each time but what happens is they get closer and closer to the actual series you want it's actually very impressive to look and see how many terms you need to get good agreement near this point you can do the same thing with e to the x and the function you want is actually very nice but unfortunately I'm cursed today so I will not do it notice that this process that I did here we line on derivatives at zero we can also do it with derivatives at other points for example at any point where we can calculate the derivative I'll come back to that in the Taylor series Taylor series is more general on the Glorin series is a Taylor series with a equals zero usually we use McLauren series because it's just easy usually it's easy to compute the function and all its derivatives at zero but sometimes you want it away from zero let me do one more example of the Glorin series who knows how much of the class but so let's say we want to go through this process for the log of one plus x okay so the reason that I'm doing one plus x is because of course the log of zero is not a nice number it's minus infinity but I want a series that will allow me to calculate the log of one plus x so that's values of x near zero the values of the log so again let's just do the McLauren series for this we start with f of x we know its derivatives or we can calculate its derivatives and we can calculate their values when we come up with this series a lot of times people if they're doing this organize this into a little chart the chart is not necessary but it helps you keep track so I'm just going to make a little chart this is f and if I want to evaluate what is f of zero well f of zero here is the log of one which is just zero so actually there isn't going to be a constant term here because it's going to start with zero so our constant term here is zero now I take the derivative the derivative of the log of one plus x the derivative of the log of x is one over x derivative of the log of one plus x is one over one plus x we use the chain rule but the derivative of one is one so that's fine and if I plug in this I guess it's not f of zero let's say half zero if I plug in zero for x I get one over one I mean one over one plus one which is a half so my coefficient term will be oh it's at zero right I was thinking I was doing it at one so it's one over one which is one so my coefficient here is one so what I have so far is that my series is zero plus one not very interesting series so far one times x maybe we can put let's just put the term in here too so the derivative of in fact let's write it this way because it's easier to keep track of the derivatives so the derivative of one plus x to the minus one is minus one plus x to the minus two which is again when I plug in zero one so this is minus one to the minus two which is still just one my coefficient here is one but here since it's the x squared term I have to divide by two because it's one over n factorial here n is two so that means that my next term and it's negative my next term in the series is minus x squared over two and if I do it again for the third derivative well this is a minus minus two times one plus x to the minus three when I plug in this is still one so this just gives me two but again it's going to be two over three factorial which is a third so my next term will be plus x cubed over three and then if I go to the next term take the derivative here I get minus two times three times one plus x to the minus four when I plug in x equals zero I just get a minus six here but again I have minus six over four factorial which is just a quarter negative so then I get minus one fourth x to the fourth yeah where did the two go? so this is two here this two so I have two but I have to divide it by three factorial because it's the coefficient of the third term so this term I always if you prefer you can write this as I just did it so I'm not going to do it this is this number divided by n factorial to go from here to here I divide by n factorial and this three in this line to go from here to here I divide in fact let's write six as two times three to go from here to here I divide by four factorial the two and the three cancel out they need to be just the four so the next term here is minus four x to the fourth and the next term if you go through it will be a fifth x to the fifth and so on and in general we can see that this is the sum of minus one to the n this is plus this is plus so I guess that has to be plus one over let's say n plus one from n equals zero to infinity something's wrong here let's start at one and then we did this before we did this also by integrating the geometric series we got exactly the same answer did we do this before maybe we did so we can calculate is this series an alternate x to the n over n and now this is can I do this right you mean on the minus one yeah four plus one I don't care because the sign's off I always do that okay so if you just look at the terms this is n equals zero four this is n equals one and it's positive so if I did minus one to the n like I wrote it should be negative so I just want to change the sign now we can do this in another way as I said so one of the nice things about math is you can do things in several ways and get the same answer because they are the same thing that helps you believe the truth of it we can also do the fact that number one plus x is a geometric series it's the geometric series with ratio minus x is one minus x plus x where minus x cubed plus x and if we just take the integral of this well then this is the log of one plus x because if you iterate the x here if we integrate one over one plus x we get the log if we integrate this geometric series what will happen well this is let me write this slightly differently minus one to the n x to the n so if I integrate that I get a constant plus so the integral here will raise the power by one minus one by the minus one to the n the x goes up by one but I have to divide by the new power and if I have been consistent here and started at zero which is what I was trying to do I would have the same thing right here is n equals zero this is x to the one and so on these are the same thing and this is also my constant plus x minus x squared over two plus x cubed over three minus x four four plus and my constant here is zero because I know that the log of zero log of one is zero so I can do these in two different ways and get the same answer in really two different ways but it's kind of two different ways but this series only converges for certain values of x what values of x does it converge for okay you should know this how would I find out do the ratio test let's do the ratio test so if I do the ratio test what I need is the ratio of the plus first term so I want to do the ratio test in that series so that means I want to look at the limit as n goes to infinity of eight n plus one or eight n which is the limit as n goes to infinity the next term will be minus one to the n plus one x to the n plus two over n plus two and then I divide by minus one to the n x to the x is on the bottom x to the n plus one or n plus one which gives me so the minus one cancel the minus one leaving the minus one but I'm taking the absolute value so I can just forget about it no limit and so the x to the n plus two is n plus x n plus one divided by n plus two is just n plus one divided by n plus two but now I take the limit as n goes to infinity this goes to one which is the absolute value of x so this convergence for absolute value of x less than one when x is one so when x is one then I have an alternating harmonic series which diverges when x is minus one the minus ones cancel out and give me the series which will wait what did I say so at x equals one the series becomes minus one to the n over n plus one which is converging because it's an alternating series with the term going to zero and when x is minus one I have n times minus one to the n plus one which is always minus one and so I have this which diverges so in fact it's for x between one and minus one but one is included so this series, this law makes sense for x to give me, I can calculate the law of two with this series but I cannot calculate the law of 2.01 with this series and I can calculate the law of any number between zero and two different ways maybe that was confusing I should have focused on only one way one way we already knew this way from last week of course maybe it didn't sink in but we already knew this way from last week last week we did integration we started with the geometric series and derived new series from it by integrating differentiating plugging in dancing around a lot of life so that's the new way and that's the old way that we knew any function will work by this way as long as you can take all the derivatives both ways work just like so your question is kind of like when do I use the integral test and when do I use the comparison test often you can use either one here you can derive the series by saying oh it's a series I know let's mess with it or just derive it from the beginning I mean your question is a good question I'm not saying what a stupid question this is a good question it's a different almost equivalent method here I took a series I know and mess with it to give a new one just like here to calculate the series for the cosine I could have done it by taking derivatives of cosine or I could just say hey this is the derivative of the sine why do any work okay alright so let me do one more example where I'm not doing it at zero the work is exactly the same way but you have to pay a little more attention so let me do let's do the series for the tangent near pi over four that will be a mess let's do the series for the cosine near pi over four so I have to remember what the cosine of pi over four is not so hard so let's get the series for cosine of x for x near pi over four so I happen to know I hope that the cosine pi over four I know this does anyone in the room know it yeah is root two over two and I will also need to know that the sine of pi over four is root two over two so in order for this series to be useful I have to know how to calculate the square root two okay so let's just start going so the cosine of pi over four I just wrote it down but I'll write it down again is root two over two so the first term in my series will be root two over two now I take the derivative let me make my chart let me make my chart so here's n here's my function if I evaluate it at pi over four and then I'll get my term so cosine of pi over four is root two over two and here that's the first term of the series so my series is root two over two plus something when I take the derivative I only get minus the sine the sine at pi over four give me a minus root two over two so then my first series will be this divided by one factorial which is just one times x minus pi over four to the one power so my next term is pi over four times how far away I am from pi over four I think this was a bad example sorry if I take the next derivative the derivative of minus the sine is the cosine which when I evaluate once again gives me this annoying root two over two but now we have root two over two divided by two factorial and then we have an x minus negative yeah there's a negative sign there don't you see it and it just goes on in this way the next term will be minus the sine again I get root two over two negative plus don't you see and so here I get root two over two times three factorial x minus pi over four cubed et cetera so my series in the end get this factor root two over two had I chosen an angle like pi over three in the switch between the cosine of pi over three and the sine of pi over three so I have that the cosine x is I can factor out this root two over two and then I get one plus x minus pi over four plus no minus x minus pi over four squared divided by two minus x minus pi over four cubed divided by three factorial plus minus minus plus plus minus oh I have a plus I have a minus here that fell in the hole thank you four factorial minus it goes plus minus minus plus plus minus et cetera so unlike at zero where the odd D when we cosine so where all of the even odd terms go away here the odd terms play in the series so the odd terms appear so what is the use of this the use of this is if I wanted to approximate the cosine of say 46 degrees then this would be a very good approximation because I just need to know the two and then these things are very small numbers that go away fast because 46 degrees minus 45 degrees is one degree convert that one degree to radians and then this converges quickly so this is a good this will converge quickly whereas this one over there will take a long time to convert because pi over four powers of pi over four don't go to zero quickly as quickly as I am of course because of this stupidness here I just hope that I won't forget so we'll have to do that next time let me remind you that there was a date for homework put it up over the weekend as well as the short web assignment