 Okay, so let's go ahead and get started even though. It looks like a few people aren't here. So what we did last time was at the end is we set up this, we set up the rate equations for this two-step reaction here where we have in the first step reactants A and B in equilibrium with an intermediate I. The forward rate constant is K1, reverse is K2. And then we have a second step where the intermediate reacts with B and produces a product C. And the rate constant for that reaction is K3. And then using this mechanism, what we did was we wrote down the rate equations that express the derivatives of each of the concentrations of the function of time and I've recapitulated them here. All right, so this is a set of four coupled first-order differential equations and what we'll do is we'll set them up in Mathematica and then generate the solutions and look at their behavior. And another thing I mentioned last time was that, you know, as written, you certainly cannot solve these equations by hand and that there's an approximation that's often used when you have something like an intermediate involved and that's called steady state approximation and it amounts to the assumption that the intermediate is never produced in large quantities and that its rate of formation is zero once it's formed. All right, so we'll have a look at that and we'll play around with the conditions and see when it is and it is not obeyed. All right, so we'll go ahead and get started then. All right, so I'll start like we did last time with a list of equations, okay, and I'm going to define concentration of A is little C big A and that's prime of T equals equals minus K1 times C A of T and times C B of T plus K2 times concentration of the intermediate as function of T, pardon, all right, I got the wrong bracket here, thank you, all right, and then for B, C B prime of T equals equals minus K1 times C A of T times C B of T plus K2 times C I of T and then minus K3 times C B of T times C I of T, comma, and then the intermediate C I prime of T equals equals K1 times C A of T times C B of T minus K2 times C I and then minus K3 times C B of T times C I of T. And then the last one, concentration of C prime equals equals K3 times C B of T times C I of T. All right, no comma. All right, let's do a comma so we can put in some boundary conditions or initial conditions. So I'll say C A of 0 equals equals C A 0, comma, C B of 0 equals equals C B 0 and then we'll take the concentration of the intermediate to be 0, which makes sense, and we'll start out with no product. So concentration of C at times 0 equals equals 0 and now that's it, that's the list. Okay, now let's try to do D solve and put the solutions into a list called solutions. So we D solve equations and the solutions we want are C A of T, C B of T, C I of T and C C of T. All right, and the independent variable is T. All right, so it looks like everything's good, so let's go ahead and try it and we see that the solution can't be generated exactly. Remember, whenever you get back what you put in, that's not a good sign. Okay, so what do we do in such a situation? Quit and D solve, yes, we don't quit. We try the numerical version. All right, so to do the numerical version, let's go ahead and copy all this stuff. We need to specify a lot more information. Okay, so let's see, first let's get rid of these guys because we're going to define them explicitly. Well, actually we can leave that here. What we'll do though is we will put in, we have to put in values for the rate constants and we have to put in values for the initial concentrations. So we'll say CA0 is equal to 1 molar and same for CB0 and then I'm going to put in that K1 is equal to 0.1 and K2 is equal to 0.1 and K3 is equal to 10. Okay, and now we can say N D solve and we have to give a range of the T over which we want the solutions. All right, so smallest K is about 0.1 so the longest time of the characteristic time of the decay is going to be around 10 so let's go ahead and make T go from 0 to 20 so we can see some action and I think we're good. So let's go ahead and see if that works and it looks good because we've got interpolating functions. All right, so now let's make a plot of all of these concentrations as a function of time. Okay, so we'll go ahead and say plot curly CA of T slash dot solutions, solution and then we can do the same for BI and C. All right, so there's what we're plotting. Okay, get rid of that comma and now we'll say we want the time to go from 0 to 20. We'll put some axes labels, just label the x-axis I guess, time. I'm going to put in a plot legend so what I'm going to do first is just make sure I have the plot legends package loaded before I get carried away here so less than, less than plot legends, backward single quote, enter. All right, so now plot legend arrow so the first thing we're plotting is the concentration of A and then B and then I and then C. Okay, and now the legend position, put it out of the way. So I'll put it at 1.0 and minus 0.2 and get rid of the shadow and get rid of the border. Okay, let's see, what else, I think that ought to do it, okay, so let's go ahead and close that off and let it rip and so there we have it. It looks like kinetics. Now, one thing that you can see is that I is pretty small and it looks like it's almost constant so already we could say that we might expect the steady state approximation would have been good here but let's have a look more closely so what I'm going to do is actually blow up I by multiplying it by 10 and I'll go ahead and indicate that in my legend, 10 times I, 10X, okay, so there you have it, all right, so now we see decrease in concentrations of our reactants and increase of the products and then you see there's a very rapid increase of the intermediate and then a little slow decay and then it definitely looks like it's fairly constant as time goes on, all right, so let's consider a different situation because it's not always like this so what we'll do is we'll go ahead and grab all this stuff, all right, and we'll change the numbers a little bit and then also we'll grab the plot here, all right. So let's change K1 to 1 and K2 to 0.2 and now K3 to 0.1, okay, so different set of rate constants and then we'll have a look and see what happened, okay, so we get our interpolating functions and now let's make the plot. Let's go ahead and take off the 10 times I and now you see, if you compare the two, got somewhat different behavior, right, so notice the reactant A here goes down quickly, had a bigger rate constant for the first step and then it actually starts to go back up as opposed to having a monotonic decrease. B is continuing with the monotonic decrease and notice that the product is being produced more slowly but most noticeably there's a big difference in the behavior of the intermediate for this set of rate constants, okay, and in particular we could see that first of all the intermediate goes up to a quite high concentration compared to all the other species and it's definitely not constant, all right, so for this particular set of parameters the exact same rate equations give quite different behavior and this would be a case where the steady state approximation would be questionable because the intermediate is substantial and not constant, okay, now when you take kinetics in physical chemistry class you'll learn about ways to sort of analyze the equations and come up with some criteria for guessing whether or not the steady state approximation would be good but here we have a nice tool where we can actually generate accurate solutions to the full set of coupled equations and then play around with the parameters to see how things change, all right, so there's one example of a fairly complicated set of kinetics equations where you can generate the solutions very easily using Mathematica, all right, so I'm going to do one more example from kinetics, all right, so let's go over to the board, any questions on that example, all right, okay, so next thing we're going to consider is that we have a reactant A reacting with something called X and that's going to give C plus Y and then we're going to have B reacting with Y to give D plus X, okay, so this is two step reaction, we'll call it the rate constant for the first step K1 and the second one K2, all right, now look what happens when I add these two together, the X's cancel and the Y's cancel, so I get the overall reaction is A plus B goes to C plus D, so in these chemical equations what do we call the species X, that's a catalyst, correct, it's something that's consumed in one step of the reaction and then regenerated in another and in principle it's supposed to speed up the rate, all right, so X is a catalyst, what about Y, Y is an intermediate, it's produced in one step and used up in another, it doesn't appear in the overall equation, all right, good, so we'll be able to see in this case how the catalyst behaves, all right, now in order to be able to solve the equations we have to first be able to write them down, so let's go ahead and write down the rate equations, so what should I write here, okay, that's, we lose A through the first step, anything else, that's it, we don't have a reverse reaction here, okay, all right, what about B, well we've done a lot of these now, by now you should know how to do it, B times Y, okay, in this case we're going to ignore the products, we need, if we're interested in just seeing how the reactants and the catalyst and the intermediate behave then we don't need to write down the equations for the products, so let's go ahead and do X and Y, okay, what about X, come on, yeah, so lost in the first one is just like A then, all right, and then gained in the second step the reactants are B and Y, and then finally Y is produced in the first step, so we have a plus, the rate constant is K1 and the reactants are A and X, then it's lost in the second step, the rate constant is K2, the reactants are B and Y, all right, so that should be good, all right, we've got equations for the concentrations of four species and we have four equations, so it should be good enough, if we wanted we could put in the products but it'll be too messy to plot, all right, so let's go ahead and get those put in and I'm going to start out like clearing a bunch of stuff because we've assigned values to CA0, CB0, CC, actually we don't need C, K1 and K2, all right, and let's go ahead and put in our equations, all right, so we have CA prime equals equals minus K1 times A times X, comma, and then B prime equals equals minus K2 times CB of T times CY of T and then X prime equals equals minus K1 times CA times CX and then plus K2 times CB times CY, comma, and then finally Y prime equals equals plus K1 times CA times CX minus K2 times CB times CY, all right, then we'll go ahead and put in some initial conditions, CA of 0 equals equals CA0, CB of 0 equals equals CB0, CX of 0 equals equals CX0 and then Y the same story, all right, so that's the hard part, so well let's just go ahead and type it in D solve equations, we're going to need a bracket, square bracket equations, which you can't type today, and then for CA of T, CB of T, CX of T and CY of T and T. Now I want to store the solutions in solution equals, all right, and let's just make sure everything looks right here, okay, it looks pretty good, so let's then go ahead and let it rip and you see once again we have a set of equations that we can't solve using D solve, all right, so let's put in some values for all the parameters, so let's start out with CA0 equals 1, CB0 equals 1, CX0 the catalyst equals 1 and we'll start out with no intermediate, CY0 equals 0 and then we'll put in K1 equals 0.5 and K2 equals 0.5 and then we'll switch to ND solve and we'll go ahead and plot this out or get the solutions for T goes from 0 to, let's do 100. All right, so let's see if that'll work, no, I think it should be okay because we're reassigning them, but I'm sorry I can't hear, all right, yes, thank you, I need to put semicolons and not commas here, that's a good one, all right, and now it should be okay, got interpolating functions, all right, so let's grab our plot command here because we're going to do more or less the same and we'll go ahead and customize it for this problem, put in X and put in Y, put in 100 for the time and then put X and Y and the legend, all right, ready, this should be good, there it is, interesting, look at the interesting dynamics here, all right, so you notice the reactants A and B, they're going down pretty fast, looks like A tails off kind of slowly, B immediately goes away practically after a little more than 10 seconds, but more interesting in this case I think is what's going on with the intermediate and also the catalyst, all right, so notice the intermediate shoots up very quickly and then undergoes a gradual slow decay that sort of mirrors the behavior of the reactant B and another thing which maybe you haven't seen before is the catalyst also has some interesting dynamics, so you see it's very rapidly consumed to a certain extent at the start of the reaction and then it slowly recovers and as you can see as time goes on it goes back to being very close to where it started which was at one, okay, so this is the behavior that we sort of assume when we talk about a catalyst, remember we said a catalyst is a species that's not formed or destroyed in the reaction, that's only true at very long times, right, you see it gets eaten up quickly at the start, all right and now that we have all this stuff typed in we can play around with it a little bit, so one thing to do is let's go ahead and make the catalyst concentration a little smaller, so let's go to 0.5, what do you suppose is going to happen? Anybody have any guesses? I'm going to change the catalyst to 0.5, well let's see, you can keep your guesses to yourself, looks the same doesn't it? The catalyst is not supposed to be the same. Is it really the same? Let's make it a bit smaller here, let's go down to 0.1, I think it might have changed a little bit, yeah, there we go, it did change, it just didn't change that much, now notice when we put in less catalyst, actually it still looks weird doesn't it, isn't it supposed to start out at 0.1, let's make this plot from 0 to 1, you know it's, oh I know what the problem is, see we have to put in CX0 here to clear and CY0, okay, sorry, all right so let's go back to 0.5, and change this to 100 again, and you can remember what it looked like, no we destroyed it, so okay, all right, go ahead and enter that, all right, so that's certainly sensibly different than it was the first time, all right, so the catalyst starts out at 0.5 where it should, and notice now it's taking longer for A to go away, and also B, the catalyst gets consumed even more in sort of a percentage wise, and then recovers later on, and let's go, let's take it down even a little bit further, let's go to say 0.1, so go ahead and enter that, and then enter that, and now notice what happens, you've really slowed the reaction down, okay, so you kind of get a feeling by playing around with these parameters how the catalyst is actually working, and under these circumstances it's going to take, well you can see that toward the end here it looks like it's getting back to close to its original value, all right, but the point is I guess for this course primarily is that you can take some seemingly complicated looking array of these differential equations, and you can fairly easily type them into Mathematica, generate a numerical solution to the differential equations, and then play around with the parameters, and in principle you can do even much more complicated coupled chemical reactions if you want, so these are sort of more complicated than you might do by hand in physical chemistry class, but not at all complicated compared to what you might do if you were say simulating the chemistry in the atmosphere, which has hundreds of reactions that are possibly worth considering simultaneously. All right, so that concludes our brief introduction to using Mathematica to solve differential equations and ending up with some chemistry related examples. Does anybody have any questions on that material? Okay, so then on the final exam you'll all get the problems cracked and the homework. So the next subject is we're going to learn how to deal with series and sums, so let me go ahead and just write down what we're going to do. Okay, so the first thing we'll do is a finite series, and that's going to be indicated by the capital sigma, and we'll have some index which we'll start at some lower value which for the purpose of example here I'll just put in zero, and then it will go up to an upper value n, and so if I was to sum a bunch of numbers that I denote a sub k, what that is is just a zero plus a one plus a two plus dot dot dot plus a sub n minus one and then plus a sub n, and then if you let n go to infinity then you get what's called an infinite series which may or may not have an actual value. Some infinite series are infinite, and so they're said to not converge while others are finite and give you a number even though you're summing over an infinite number of terms. That's called a convergent series. Okay, so we'll start out by just seeing how to enter in series and we'll see a few examples. Okay, so the way you sum up a series is you say sum, okay, and then you put it in what it is you want to sum, so I'm going to say give me the sum of k squared, and then I have to give a range of k, all right? So I can say k goes from 1 to 20, okay? So what this should be is the sum of 1 squared plus 2 squared plus 3 squared dot dot dot up to 20 squared, all right? So let's see what we get, 2,870. I could have also done the same thing in a different way. I could define a function f of k colon equals k squared, and then I could say sum f of k, k goes from 1 to 20, and I have the wrong kind of bracket here, all right? And we should get the same answer. Now, if you want a numerical result, you can do n percent, here it just puts a decimal point, or you could just do directly, you could say instead of saying sum, you could say n sum, and then you get the approximate, okay? Let's do another one, f of k colon equals, sign of k over k, and then we'll do a sum. Just make this one go up to 10, and notice, you get the exact representation. If you want the numerical value of that, you can just do n sum, okay? So that turns out to be 1.12, all right? You can also sum over a range of values that has a variable as one or more of the indices. So for example, there's a series called the geometric series, which is a partial sum of, let's see it here, sum x to the power k, and we could say k goes from 0 to n, notice n here, which is just a variable at this point. Okay, so you sum that, and it's very interesting that Mathematica recognizes that that sum, the geometric series, has this closed form, x to the 1 plus n minus 1 over x minus 1, it's a famous series. All right, another way to do sums is you can come over here to the pallet and grab the sum symbol here, okay? And then you can put in k, 0, oops, 0, tab n, tab x, tab k, and you should get the same answer, but it looks nice. And you could do sum of k, tab 1, tab 10, we could do our sign, oops, I didn't mean to do that, k, tab 1, tab 10, tab, sign of k, tab k, so that'll give the same result that we got before, all right? So if you like to use the pallet, there you have it. That's how you do it for sums, be careful though, as usual. All right, so now let's do some infinite series, all right? So sometimes it works, sometimes it doesn't, and those who have had the appropriate calculus class know how this works, sometimes you get very interesting results when you do infinite series. So one interesting one is we could give the sum of 1 over k factorial from, k goes from 0 to infinity. Anybody know that one? Sort of, he said natural log e. So that's the sum representation of the number e, interesting. All right, let's do sum of 1 over, 2 to the power k, k goes from 0 to infinity, bracket. Anybody remember that one? Interesting. All right, let's try one with the pallet here. So let's do sum k goes from 0 to infinity, tab. This is going to be 1 over, and then we're going to have quantity 2k plus 1 squared, all right? So we'll do this, and then we'll do this, 2k plus 1, tab squared. Anybody know that one? Does it converge? Let's see, pi squared over 8 percent, which is about one and a quarter. Interesting how these sums converge, isn't it? All right, let's do, let's try this one, sum 1 over k, k goes from 1 to infinity. That one doesn't converge. Interesting, huh? Because you think, gee, as k gets bigger, I'm adding a smaller and smaller number, but still it doesn't converge. All right, let's do another version of the geometric series, all right? So we're going to say sum of a number times x to the power n, all right, and we'll say n goes from 0 to infinity, all right? Let's try that, get a over 1 minus x. Now, this is an interesting one because let's suppose that we say, I want to say that x is greater than 0, or greater than 1, sorry, so that one doesn't converge, but when x is less than 1, it does, all right? So that's a case where you have conditional convergence depending on what the value of x is, all right? So here we get a nice closed form for the geometric series, whereas it doesn't converge when x is greater than 1. All right, so that's some simple summations, both finite and infinite, a few examples. And now what I want to do is talk about expansions of functions in series. This is a pretty cool thing that you can do with Mathematica, so I'm going to go back over to the board. And some of you have seen this before, and probably some have not, but in any case, these are used quite frequently in physical chemistry, and when you learn them, or when you need them, I'm sure your instructor will tell you how they work, but this will just be a quick crash course, okay? So there's different types of ways to expand functions, okay? So one is what's called a power series. If I have a function f of x, its power series representation would be given by a constant, which we'll call a zero, and then a coefficient times x plus another coefficient times x squared plus dot, dot, dot, so cubed, fourth, et cetera, okay? Now, when you have a function that has continuous derivatives of all orders, so that's the assumption here, you can expand your function around a point, okay? And you can expand it around that point in terms of a sum over its derivatives evaluated at that point. So for example, one form is you can say that the function is equal to its value at the origin plus its derivative at the origin times x plus its second derivative evaluated at the origin divided by 2 times x squared plus third derivative divided by 3 factorial, so this is factorial, times x cubed plus dot, dot, dot, okay? So you're adding more information. You have the, here, this is sort of a line. This is a line form, and then you're adding in some curvature with the quadratic, and then you add an additional flexibility into the function as you add more and more terms, and you get a more and more accurate representation of the function, all right? Now, another version is you could, instead of expanding the function around the origin, x equals 0, you can expand it around a point, which I'll call h. So we would say f of x plus h is equal to f at the value h plus f prime at the value h times, x minus h plus the second derivative evaluated at h times x minus h squared, et cetera, et cetera, okay? And this is what's called a Taylor series, and this one here is what's called Maclaurin, which is a special case of the Taylor series where the point that you're expanding around is equal to the origin, okay? How many people have had some experience with these representations? A few, yeah, okay. All right, so you know how this works. Well, in any case, we're going to see how it works because we're going to play around with Mathematica, and it's kind of interesting to see how these expansions actually build up the function, all right? So we'll play with that, and then we'll, later on next week, I guess, we'll do some examples from chemistry where the series and sums show up, so it'll be a preview of the future, okay? So let's go ahead and generate series. So the first thing we'll do is we'll just show that Mathematica knows how to do Taylor series, okay? And I'm going to clear f, which I've used earlier, and then I'm just going to generate a generic Taylor series by saying give me a series representation of the function f of x, and x is the independent variable. I want to expand around the point x equals h, and I want it to third order. That's what all this means, okay? Function to be expanded, the independent variable, the point around which you want to expand, and the number of terms, all right? So let's see what we get. So you see what we get is, oh, right, sorry. So we get exactly what's written on the board over there, f of h, f prime times x minus h, and then second derivative times x minus h squared divided by 2, and then we get the third order term here, all right? So 3 gives us out to the third order, third power. And then there's this thing plus oh, x minus h to the fourth. The interpretation of that is that the next term, the oh here means of the order of. So it says the next term, which we're neglecting, is of the order of x minus h to the fourth power. So it's proportional to that, all right? Now, if you wanted to use this expression in a calculation, which you could do, you could set it equal to something. This thing would present a problem, because that can't be evaluated, all right? So the way you can get rid of that is you can put a little wrapper around the command, which is called normal, and that's kind of nice because what that does is it deletes that last term, all right? So now, if you actually wanted to use this, you could. Okay, let's do a couple more examples. So now what we'll do is we'll expand e to the x in a Taylor series, all right? So that means around the origin. And I'll do it using normal. So I say normal series. I want to expand e to the x, x, zero, I'm doing it around the origin, and then to tenth order, okay? Now, if I do that, oops, I find out that I'm missing a bracket, and then I see the series, okay? So that's the series representation of e to the x out to order 10, all right? Now, the next thing I want to do is to show you how you can have a look and see how these different terms sort of add in accuracy to representing e to the x, all right? So what we're going to do is I'll go ahead and assign this to a function, okay? So I'm going to say f of x underscore colon equals, all right? And if I do question mark f, I see there it is, all right? And now if I want, I could actually evaluate it. So for example, let's do e to the 1 power, all right? So we do that, oops, let's do it this way. f of x, I don't, I want to use a replacement rule. So x, shoot, x arrow 1, okay? So that will evaluate it for 1, okay? So there's the exact, or the exact calculation based on 10 terms of the series. We could say n percent comma 10 places, all right? And now let's compare that to the exact result. So this is an approximation based on the 10th order series. So let's evaluate n of e to the 1 to 10 places, all right? And now you can see just how accurate that series is, right? These numbers are coincident out to this place here. All right, so that 10th order Taylor series was quite accurate. And now what we'll do is we'll have a look and see how that accuracy comes in as you add more and more terms. All right, so we're going to generate a bunch of curves. Each curve is going to be a subsequent order of the Taylor series. And then we'll plot all the curves along with the exact result for e to the x. And you'll be able to see how as you add additional terms, you get a more accurate representation. All right, so what I'm going to do then is I'll say I'm going to store some data into a table. So I'll say series data equals table. And then I'll just grab this thing here and put it in there. Okay, and I'm just going to do six terms so things don't get too ugly here. All right, and then we'll have, no, this is going to be, sorry, this is going to be x goes from zero to i. And then i goes from zero to, no, sorry, one to six. All right, so what's this going to do? All right, so i is going to start out at one. And so what this will do, the first entry in the table is going to be the series with just one term. Okay, and then i is going to be two. And we'll have two terms, three, four, five, six. The last one will be six terms. All right, and it looks like I have all the brackets that I need, so I'll go ahead and put in a semicolon. Well, let's actually see what that looks like. All right, so there you have it. First, one term, two, et cetera. All right, so we have all the series. And now what we'll do is we'll make a plot of those. Series plot equals plot, series data. You need a bracket, sorry. And I'm going to have it go from x goes from zero to two. Okay, and I'll put a semicolon. And now I'm going to generate another plot that has the exact representation of e to the x. And then we'll just plot them on the same plot using the show command. So I'll call this exact plot equals plot e to the x. X goes from zero to two. And I'm going to make it a thick black line so that we can tell which one is the exact plot and then the other ones will be colored. Okay? So the way to do that is say plot style, arrow, curly bracket, gray level, zero, and thickness, 0.005. Okay, curly bracket, square bracket, semicolon. And then we'll do show, exact plot, and series plot. All right, let's see what we get. Okay, so there you have it. So hopefully this will help you to understand how this series works. All right, so the thick black line, that's just plain old e to the x, exactly calculated. And this blue one, that's the first order expansion which is just a line, it's one plus x. And you see that only really close to x equals zero is the line a good representation. All right? But if you want to have more accuracy as you go out further and further away from zero, you need more and more terms. Okay? And you can see the effect of adding those additional terms. So here's the second order term which has some parabola in it. And here's the cubic. And you can see the fourth already gets you out pretty far. And then fifth and the sixth is hardly distinguishable from the exact curve over this range. All right, so this shows you how the Taylor series actually works and teaches you something important about it which is that if you want more and more accuracy further and further away from the point around which you're expanding, then you need to add more terms. Okay? All right. Couple of other things to point out. So let's look at some trigonometric series. All right, so let's do series of cosine. And we'll do this for a Taylor series, zero, and we'll put in eight terms or up to eighth order. All right? So what's interesting about this one? Correct. So what he said was that all the odd powers are zero. Interesting. Cosine is what's referred to as an even function and it has only even orders in its Taylor series. Does everybody know what an even and an odd function are? Even function is a function that's symmetric about the origin. So f of minus x equals f of x. And an odd function is anti-symmetric, f of minus x equals minus f of x. So sine is an odd function. So let's see, let's do the same one for sine. What do you think we're going to get here? Odd terms. Does that make sense? If I have any odd terms, if I have odd terms, right, then minus x cubed is equal to x cubed, right? So that's actually, that's what I said that in a funny way. All right, if you put in one into x cubed, you get one. If you put in minus one, you get minus one, right? Whereas if you put one into x squared, you get one. And you put minus one into x squared, you get one, okay? So an even function can't have odd terms and vice versa. Odd order terms. All right, so let's do another one where we'll do the expansion of log. So we'll do log of one plus x. And we'll expand that around a point x equals a and do five terms, all right? So there you get a nice handsome looking series that's got the a in there. Okay, so let's see here. Trying to decide whether or not we should quit here because the next subject requires a bit of explanation. So actually I think that we'll go ahead and finish a little early today so you can study up for the exam. Give you an extra 10 minutes to study up for the exam. And then the next subject that we'll start on Monday, is Fourier series. And just to give you a quick preview, Fourier series are series that involve sums of sines and cosines. And those are very useful for representing periodic functions. And in particular, they're very useful in doing things like signal processing, all right? So that's what we'll do on Monday. So see you on Thursday for your exam. Have a nice couple of days.