1. Math 104 - Calculus I August 11 (but first, a quick review…)

2. Power series Last week's project was to try and sum series using your calculator or computer. The answers correct to ten decimal places are: Sum((-1)^n/(2*n+1),n=0..infinity) = evalf(sum((-1)^n/(2*n+1),n=0.. infinity)); Sum(1/factorial(n),n=0..infinity)=evalf(sum(1/factorial(n),n=0..infinity) );

3. Power series (cont.) Sum(1/n^2,n=1..infinity)=evalf(sum(1/n^2,n=1..infinity)); Sum((-1)^(n+1)/n,n=1..infinity)=evalf(sum((-1)^ (n+1)/n,n=1..infinity)); We can recognize these numbers as

4. Two directions: 1. Given a number, come up with a series that has the number as its sum, so we can use it to get approximations. 2. Develop an extensive vocabulary of "known" series, so we can recognize "familiar" series more often.

5. Geometric series revisited

6. r as a variable Changing our point of view for a minute (or a week, or a lifetime), let's think of r as a variable. We change its name to x to emphasize the point: So the series defines a function (at least for certain values of x).

7. Watch out... We can identify the geometric series when we see it, we can calculate the function it represents and go back and forth between function values and specific series. We must be careful, though, to avoid substituting values of x that are not allowed, lest we get nonsensical statements like

8. Power series If you look at the geometric series as a function, it looks rather like a polynomial, but of infinite degree. Polynomials are important in mathematics for many reasons among which are: 1. Simplicity -- they are easy to express, to add, subtract, multiply, and occasionally divide 2. Closure -- they stay polynomials when they are added, subtracted and multiplied. 3. Calculus -- they stay polynomials when they are differentiated or integrated

9. Infinite polynomials So, we'll think of power series as "infinite polynomials", and write

10. Three (or 4) questions arise... 1. Given a function (other than ), can it be expressed as a power series? If so, how? 2. For what values of x is a power series representation valid? (This is a two part question -- if we start with a function f(x) and form "its" power series, then (a) For which values of x does the series converge? (b) For which values of x does the series converge to f(x) ? [There's also the question of "how fast".]

11. continued 3. Given a series, can we tell what function it came from? 4. What is all this good for? As it turns out, the questions in order of difficulty, are 1, 2(a), 2(b) and 3. So we start with question 1:

12. The power series of a function of f(x) Suppose the function f(x) has the power series Q. How can we calculate the coefficients a from a knowledge of f(x)? A. One at a time -- differentiate and plug in x=0! i

13. Take note...

14. Continuing in this way...

15. Example Suppose we know, for the function f, that f(0)=1 and f ' = f. Then f '' = f ', f ''' = f '' etc... So f '(0) = f ''(0) = f '''(0) =... = 1. From the properties of f we know on the one hand that So we get that...

16. Convergence of power series Before we get too excited about finding series, let's make sure that, at the very least, the series converge. Next week, we'll deal with the question of whether they converge to the function we expect. But for now, we'll assume that if they converge, they converge to the function they "came from". (Strictly speaking, this is not always true -- but it is true for a large class of functions, which includes nearly all the ones encountered in basic science and mathematics. This fact was not fully appreciated until the early part of the twentieth century.) Fortunately, most of the question of whether power series converge is answered fairly directly by the ratio test.

17. Recall that... for a series of constants, we have that the series converges (absolutely) if the the limit of the absolute value of is less than one, diverges if the limit is greater than one, and the test is indeterminate if the limit equals one. To use the ratio test on power series, just leave the x there and calculate the limit for each value of x. This will give an inequality that x must satisfy in order for the series to converge.

18. For the series we just calculated...

19. Your turn... Calculate the series for the function sin(x) and determine for which x the series converges.

20. Here’s a more interesting example

21. What remains...

22. Final Conclusion

23. OK, your turn... A. -1 < x < 1 B. -2 < x < 2 C. 1/2 < x < 1/2 D. -2 < x < 2 E. -1/2 < x < 1/2

24. One more... A. -1 < x < 1 B. -1 < x < 1 C. 1 < x < 1 D. -1 < x < 1 E. 0 < x < 1

25. From these examples,...it should be apparent that power series converge for values of x in an interval that is centered at zero, i.e., an interval of the form [-a, a], (-a, a], [-a, a) or (-a, a) (where a might be either zero or infinity). The interval is called the interval of convergence and the number a is called the radius of convergence.

26. Let’s go back To finding series of functions:

27. The other way

28. Try this... Take the derivative of the series for sin(x) to get

29. Integrate both sides of the geometric series from 0 to x to get:

30. Negate both sides and replace x by (-x) everywhere to get:

31. Start from the geometric series again... And substitute x for x everywhere it appears to get 2

32. A challenge to think about... How to get the other one from previously

33. 1. Limits: Series give a good idea of the behavior of functions in the neighborhood of 0: We know for other reasons that We could do this by series: Application of Series

34. This can be used on complicated limits... Calculate the limit: A. 0 B. 1/6 C. 1 D. 1/12 E. does not exist

35. Application of series (continued) 2. Approximate evaluation of integrals: Many integrals that cannot be evaluated in closed form (i.e., for which no elementary anti-derivative exists) can be approximated using series (and we can even estimate how far off the approximations are). Example: Calculate to the nearest 0.001.

36. We begin by...

37. According to Maple... The last series is an alternating series with decreasing terms. We need to find the first one that is less than 0.0005 to ensure that the error will be less than 0.001. According to Maple: evalf(1/(7*factorial(3))), evalf(1/(9*factorial(4))),evalf( 1/(11*factorial(5))); evalf(1/(13*factorial(6)));.02380952381,.004629629630,.0007575757576.0001068376068

38. Keep going... So it's enough to go out to the 5! term. We do this as follows: Sum((-1)^n/((2*n+1)*factorial(n)),n=0..5) = sum((-1)^n/((2*n+1) *factorial(n)),n=0..5); evalf(%);.7467291967=.7467291967

39. and finally... So we get that to the nearest thousandth. Again, according to Maple, the actual answer (to 10 places) isevalf(int(exp(-x^2),x=0..1));.74669241330

40. Try this... Sum the first four nonzero terms to approximate A. 0.7635 B. 0.5637 C. 0.3567 D. 0.6357 E. 0.6735

41. Application of series (cont.) 3. Differential equations : Another important application of series is to find "solutions" to differential equations (when other methods fail). Earlier, we found the series for based on the differential equation it satisfies (y ' = y ), together with the initial condition y(0)=1.

42. Airy functions The equation y '' + xy = 0 has no "elementary" solution (the solutions of this equation are called "Airy functions", named after a famous British Royal Astronomer). But we can find series for the solution that satisfies y(0)=1, y '(0)=0, as follows: From the initial conditions, we can assume that the series for y(x) begins:

43. Airy (continued) Then the series for series:

44. So... We work our way from left to right -- since y’’+ xy is supposed to be zero, every coefficient must be zero. Therefore:

45. So the series for y(x) begins as follows...

46. What is the denominator of the x term of this series? A. 720 B. 1080 C. 1440 D. 4440 E. 12960 9

47. Application of series (cont.) 4. Algebraic equations depending on a parameter ("regular perturbations"). Consider the equation. We could calculate the solutions of this equation using the quadratic formula, but it will be instructive to think of the two solutions (x) as functions of the parameter a. We will find the power series of one of these functions. Keep in mind that we are thinking of x as a function of a (and not the other way around)! Write x = f(a). First, if a=0, there are two roots, x = 2 and x = -2. We'll calculate the series for the larger root, so f(0)=2.

48. Work it out...

49. Right to left

50. Graph To see how good the approximation is, we plot the "actual" solution (in blue) and our approximation (in red) on the same graph

51. And…. What is the beginning of the series for the other root?

52. A max/min problem During World War II, the blood of thousands of recruits had to be tested for various diseases. The incidence of the diseases was quite low (less than one per hundred). Assume that the probability that a random person has the disease is p (for some p between 0 and 1, but much closer to 0 than to 1). Let q = 1 - p be the probability that the person does not have the disease. Because the tests were costly, someone had the bright idea to pool several (x) samples at a time. If the pooled sample tested negative, then all x samples were negative. But if the pooled sample tested positive, then each individual sample must be tested again.

53. max/min

55. Rewrite

56. Undaunted... we try expanding the terms in powers of p. We know that ln(1-p) = -p +... and If we substitute this much into the equation, we get: The solution of this is (which is at least consistent with the limit!). What we have here is an example of an asymptotic series, or, a series of powers other than whole numbers. We can calculate more coefficients by expanding the equation in higher powers of p.

57. Good night…. Don’t Forget: - office hours - hand in homework - problems of the day See you Monday!