课程资料 18-01-fall-2006 18-01-fall-2006 contents lecture-notes lec38

1、MIT OpenCourseWare 18.01 Single Variable CalculusFall 2006For information about citing these materials or our Terms of Use, visit: /terms.Lecture 3818.01 Fall 2006Lecture 38: Final ReviewReview: Dierentiating and Integrating Series.If f (x) =anxn, thenn=0anxn+1 n +

2、1nanxn1f (x) =andf(x)dx = C +n=1n=0Example 1: Normal (or Gaussian) Distribution. x + dt x0(t2)2 2!(t2)3 3!2etdt =1 t2 +0 x t4 t6t8 24! . dt1 t+2!3!+=0x3 +1 x51 x7= x + .32! 53! 7 xEven thoughet2 dt isnt an elementary function, we can still compute it. Elementary functionsare still a little bit bette

3、r, though. For example:0sin x = x x3 + x5 = sin=(/2)3 + (/2)53!5!223!5!But to compute sin(/2) numerically is a waste of time. We know that the sum if something very simple, namely,sin= 12Its not obvious from the series expansion that sin x deals with angles. Series are sometimes com plicated and uni

4、ntuitive.Nevertheless, we can read this formula backwards to nd a formula forStart with sin = 1. .22Then,1 dx= sin1 sin0 = 0 = 21111= sinx21 x200We want to nd the series expansion for (1 x2)1/2, but lets tackle a simpler case rst: 1 1 21111(1 + u)1/2 = 1 + 2u + 2 2 11 2 2 2 1 2 1 2 3u2 +u3 +11 3 21

5、3 5 u3= 1 u +u + 22 42 4 6Notice the pattern: odd numbers go on the top, even numbers go on the bottom, and the signsalternate.1Lecture 3818.01 Fall 2006Now, let u = x2.(1 x2)1/2 = 1 + 1x2 + 1 3x4 + 1 3 5 x6+ 22 42 4 61 3 x51 3 5 x71 x3(1 x2)1/2 dx = C +x +2 4 52 4 6 72 3 1=(1 x2)1/2dx = 1 +0 1 211

6、311 3 51+232 42 4 657Heres a hard (optional) extra credit problem: why does this series converge?LHpitals rule to nd out how quickly the terms decrease.Hint: useThe Final ExamHeres another attempt to clarify the concept of weighted averages.Weighted AverageA weighted average of some function, f , is

7、 dened as:b w(x)f (x) dxa bAverage(f) =a w(x) dx bHere,w(x) dx is the total, and w(x) is the weighting function.aExample: taken from a past problem set.You get $t if a certain particle decays in t seconds. How much should you pay to play? given that the likelihood that the particle has not decayed (

8、the weighting function) is:You werew(x) = ektRemember, 1kekt dt =0The payo isf(t) = tThe expected (or average) payo is tekt dt f (t)w(t) dt00= ekt dtw(t) dt00 = ktekt dt =(kt)ekt dt00Do the change of variable:u = kt and du = k dt2Lecture 3818.01 Fall 2006dukAverage =ueu0 On a previous problem set, y

9、ou evaluated this using integration by parts:ueudu = 1.0 du1Average =ueu=kk0On the problem set, we calculated the half-life (H) for Polonium120 was (131)(24)(60)2 seconds. We also found thatk = ln 2HTherefore, the expected payo is1H=kln 2econds.where H is the half-life of the particleNow, youre all

10、probably wondering: who on earth bets on particle decays?In truth, no one does. There is, however, a very similar problem that is useful in the real world. There is something called an annuity, which is basically a retirement pension. You can buy an annuity, and then get paid a certain amount every

11、month once you retire. Once you die, the annuity payments stop.You (and the people paying you) naturally care about how much money you can expect to get over the course of your retirement. In this case, f (t) = t represents how much money you end up with, and w(t) = ekt represents how likely your ar

12、e to be alive after t years.What if you want a 2-life annuity? Then, you need multiple integrals, which you will learn about in multivariable calculus (18.02).Our rst goal in this class was to be able to dierentiate anything. In multivariable calculus, you will learn about another chain rule. That chain rule will unify the (single-variable) chain rule, the product rule, the quotient rule, and implicit