MATH CALCULUS II SPRING Scientia Imperii Decus et Tutamen 1

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1 MATH CALCULUS II SPRING 2019 Scientia Imperii Decus et Tutamen 1 Robert R. Kallman University of North Texas Department of Mathematics 1155 Union Circle Denton, Texas office telephone: office fax: office robert.kallman@unt.edu January 4, Taken from the coat of arms of Imperial College London.

2 SPRING 2019 COURSE: MATH , CALCULUS II. PREREQUISITES: MATH CLASS MEETS: Monday, Wednesday, Friday 10:00 a.m. - 10:50 a.m., CURY 211. FINAL EXAM DATE AND TIME: The final is scheduled for Saturday, May 4, 2019 in CURY 211, 8:00 a.m. - 10:00 a.m. TEXT: James Stewart, Calculus, Eighth Edition, Cengage Learning, Boston, MA 02210, copyright 2016, 2012, ISBN INSTRUCTOR: Robert R. Kallman, 315 GAB [office], [office telephone], [fax], robert.kallman@unt.edu [ ] OFFICE HOURS: Monday, Wednesday, Friday, 8:00 a.m. - 9:30 a.m. & 11:00 a.m. - 12:30 pm ATTENDANCE POLICY: Mandatory. Specifically for TAMS students: if you are absent for any reason, you are required to file an absence report with the appropriate official in the TAMS Academic Office. ELECTRONIC DEVICES: No electronic devices of any sort are to be on let alone used during the class. Repeated flouting of this will result in a grade penalty. HOMEWORK: Homework will be assigned and some designated subset of it will be graded. The designated homework assigned on Monday, Wednesday and Friday of one week will be due at the beginning of class on the Wednesday of the following week. Late homeworks will not be accepted under any circumstances. This applies even if the student is ill - give your homework to a classmate to hand in on time or convert your homework into a *.pdf file and to me with a time stamp earlier than the beginning of class. Each homework problem will receive a grade of 0, 1, or 2 points. Failure to turn in a homework set on time will result in a grade of -1 for that homework set. ACADEMIC INTEGRITY: There is no reason why TAMS students should not demonstrate complete academic integrity at all times, particularly on tests. Any transgression of this will result in a grade of zero on the test and a grade of F for the course. Consistent with this policy the instructor will retain xerox copies of a random sampling of all tests. 1

3 GRADING POLICY: Grades will be based on the total number of points accrued from the assigned graded homeworks, from two in class one hour examinations (5 problems plus 1 bonus question), given circa in late February and late April, and from an in class 120 minute final (8 problems plus 1 bonus question). The number of points per test and final problem will normally be 10. There is no excuse for missing a test and no makeup tests will be given under any circumstances. A student missing a test will receive a grade of -1 on that test. If a student is unavoidably absent from a test and makes arrangements with the instructor well before the test date, then the grade assigned to the missing test will be prorated by the student s performance on the final examination minus 10 points. It is difficult a priori to determine the precise break points for the final grades. However, the golden rule in determining the final assigned grade is that if the number of points earned by person A is to the number of points earned by person B, then person A has a grade which is to the grade of person B. The only possible exception is that you must take the final examination and receive a passing grade on the final in order to get a grade greater than F. Just as in last semester, students can obtain a real time good first approximation to their progress/status in this class by doing the following calculation. Let S = 0.4( A B ) + 0.6( C D ), where A = number of homework points earned to date, B = maximum number of homework points possible to date, C = number of test and final points earned to date and D = maximum number of test points possible to date (not counting bonus points). Note that 0.0 S 1.0. Then the grade in this class at this instant is: A if S 0.9; B if 0.8 S < 0.9; C if 0.7 S < 0.8; D if 0.6 S < 0.7; F if S < 0.6. TOPICS: The topics to be covered can be found in most of Chapter , 6.8, , 7.8, 8.1, 8.2, 9.3, 9.5 and an independent study of Taylor s formula with remainder. It is a rather ambitious goal to cover these topics in some depth. This will require considerable work on the part of the students and the instructor. Some supplementary notes will be handed out. APPROXIMATE ITINERARY: The following is a first attempt, very rough approximation to what our schedule will be. This will perhaps be dynamically reconfigured as the semester progresses since it is of course impossible to make such a schedule with hard-andfast rules. Test #1, Friday, February 22, 2019, covering , 6.8 and 7.1. Test #2, Friday, April 12, 2019, covering , 7.8, 8.1, 8.2, 9.3, 9.5, Selected portions of Chapter 11, especially an independent study of Taylor s Formula with the integral form of the remainder. 2

4 Final, Saturday, May 4, 2019, cumulative. ASK QUESTIONS in class so that we may all benefit. If you need help, it is your responsibility to seek me out. See me during my office hours. Empirical evidence suggests that there is a strong correlation between the amount of work done by the student and his/her final grade. STUDENTS WITH DISABILITIES: It is the responsibility of students with certified disabilities to provide the instructor with appropriate documentation from the Dean of Students Office. STUDENT BEHAVIOR IN THE CLASSROOM: The Powers That Be have strongly suggested that students be given the following statement: Student behavior that interferes with an instructor s ability to conduct a class or other students opportunity to learn is unacceptable and disruptive and will not be tolerated in any instructional forum at UNT. Students engaging in unacceptable behavior will be directed to leave the classroom and the instructor may refer the student to the Center for Student Rights and Responsibilities to consider whether the student s conduct violated the Code of Student Conduct. The university s expectations for student conduct apply to all instructional forums, including university and electronic classroom, labs, discussion groups, field trips, etc. The Code of Student Conduct can be found at In other words, cause trouble in the classroom and you will probably be cast into the Darkness and sent to the KGB. 3

5 How to Study for This Class Attend every class. Pay attention in class, take careful notes, and ask questions if needed. The evening of every class go over your classroom notes, list topics on which you have questions or need clarification, read the relevant section of the textbook, do the assigned homework to be graded, look over the additional homeworks to verify that you understand how to do them and make note of those additional homeworks that you do not understand to ask about them during the next class. It is important that you put a great deal of effort into the homework, both those to be turned in and those that are less formally assigned. One becomes adroit at any human activity - e.g., hitting a fast ball, throwing a slider, making foul shots or jump shots, or driving off a tee - only with a great deal of practice. The same applies to calculus. Do not waste your time memorizing endless lists of derivatives and antiderivatives. This in fact is counterproductive. Instead, know a few basic computational techniques (e.g., product rule for differentiation, chain rule for differentiation, sin = cos, etc.), and try to understand the big picture and concepts involved in problem solving. All of the problems enountered in this class should be first approached by asking oneself what is a reasonable way to proceed. Then given the proper path or direction, you can then solve the problems by small, logical steps that inevitably lead one to the final solution. 4

6 Recall essential facts from 1710 and before: f or Df given by f f(x+h) f(x) (x) = lim h 0. h (f + g) = f + g, (af) = af, (fg) = f g + fg (Product Rule), (f/g) = (f g fg )/g 2 (Quotient Rule), h = f g = h = (f g)g (Chain Rule), (x n ) = nx n 1, derivatives of polynomials and rational functions, (f n ) = nf n 1 f, addition formulas for sin and cos, double angle formulas, and half angle formulas, cos 2 (x)+sin 2 (x) = 1, sin = cos, cos = sin, tan = sec 2, sec (x) = sec(x) tan(x), Fundamental Theorems of Calculus, continuous functions assume maximum and minimum on closed bounded intervals, local maxima and minima on open intervals, derivatives at local maxima and minima on open intervals, Rolle s Theorem, Mean Value Theorem, f = 0 = f is a constant, f > 0 = f, f < 0 = f, F 1 = F 2 = F 1 = F 2 +C, one-to-one onto functions and their inverses and derivatives. Volumes of solids of rotation and arc length formula. The surface area of a cone and the surface area generated by a surface of revolution. Volume and surface area of a sphere and spherical caps and a circumscribing cylinder, recalling profound work of Archimedes. L Hôpital s Rule. 5

7 Theorem 1 (Darboux). Let a, b R and let f : [a, b] R be differentiable. If λ R is between (Df)(a) and (Df)(b), then there exists c [a, b] such that (Df)(c) = λ, i.e., even though Df need not be continuous, it does have the intermediate value property. If (Df)(a) = (Df)(b), there is nothing to do. Otherwise we may suppose that (Df)(a) < (Df)(b), for if not, replace f with f and λ with λ. Let g(t) = f(t) λt. Then (Dg)(a) < 0 and (Dg)(b) > 0, so that g(t 1 ) < g(a) for some t 1 a, b and g(t 2 ) < g(b) for some t 2 a, b. Hence, g assumes its minimum on [a, b] at some point c a, b. Hence, (Dg)(c) = 0 = (Df)(c) = λ. Definition 2. R # = [, + ]. Theorem 3 (L Hôpital s Rule). Let a, b R #, f, g : a, b R differentiable, and (Dg)(x) 0 for all x a, b. Suppose that (Df)(x) A R # as x a. If either f(x) 0 and g(x) 0 as x a or if (Dg)(x) g(x) ± as x a, then f(x) A as x a. The analogous statement is true if x b. g(x) Dg 0 on a, b and Darboux s Theorem 1 implies either Dg > 0 or Dg < 0 on a, b = g is strictly monotone on a, b. In particular, if g(u) 0 as u a, then g(u) 0 for all u a, b. Suppose that A < +. Choose q R with A < q. There exists c a, b such that x a, c = (Df)(x) (Dg)(x) < q. If a < x < y < c, then Cauchy s Mean Value Theorem shows that there is t x, y such that f(x) f(y) = (Df)(t) g(x) g(y) (Dg)(t) < q. Suppose that f(u) 0 and g(u) 0 as u a. Letting x a we have that f(y) g(y) q for any y a, c. Next, suppose that g(u) + as u a. Keeping y fixed, we can choose a point c 1 a, y such that g(x) > g(y) and g(x) > 0 for x a, c 1. Now multiplying by g(x) g(y), we obtain g(x) f(x) < q q g(y) g(x) g(x) g(x) for all a < x < c 1 < y. Let x a we have that there is some c 2 a, c 1 such that f(x) g(x) < q for a < x < c 2. A similar argument applies in case g(u) as u a. In every case, for any A < q there is some a < c 2 < b such that a < x < c 2 = f(x) f(x) g(x) x a f(x) g(x) x a A. A very similar g(x) f(x) argument also shows that < A + = A lim inf x a. Of course a virtually g(x) identical proof shows that the analogous statements are true if x b. 6

8 Warning. Apply L Hôpital s Rule, like you would any theorem, only to situations where the hypotheses hold. 7

9 Section 6.1. Inverse Functions. One-to-one functions, examples, horizontal line test, inverses, a function has an inverse if and only if it is one-to-one, finding inverses, examples, derviatives of inverse functions, the derivative rule for inverses, examples. Derivatives of inverse functions. Use smooth tangent line arguments to show that derivatives of f 1 exit. Then (f 1 (f(x)) = x = (f 1 ) (f(x))f (x) = 1 = (f 1 ) (f(x)) = 1/f (x) = 1/f (f 1 (f(x))) = (f 1 ) (y) = 1/f (f 1 (y)) whenever the denominator of the last term is nonzero. Alternatively, and perhaps simpler, f(f 1 (y)) = y = f (f 1 (y))(f 1 ) (y) = 1 = (f 1 ) (y) = 1/f (f 1 (y)) whenever the denominator is nonzero. Example 4. f(x) = x 2 for x > 0 and f 1 (y) = y for y > 0. f (x) = 2x and therefore (f 1 ) (y) = 1/f (f 1 (y)) = 1/f ( y) = 1/2 y. Example 5 f(x) = x 3 2 = f (x) = 3x 2. Find (f 1 ) (6). Note that f(2) = 6 = f 1 (6) = 2. (f 1 ) (6) = 1/f (f 1 (6)) 1/f (2) = 1/12. Exercise 1. Look over Section 6.1 and the asssociated exercises. You should should be able to do #17 - #28, #33 - #46, #49, #50, pp Graded: #22, #27, #41, #49, #50, pp No graphing. These homeworks are due at the beginning of class on Wednesday, January 23, Remember that homeworks make up circa 40% of your grade. 8

10 Section 6.2* & 6.3*. The Natural Logarithmic and National Exponential Functions. Logarithms and their practical importance. Definition 6. If x > 0, define log(x) log e (x) ln(x) = x 1 dw/w. Theorem 7. ln(1) = 0, x > 1 = ln(x) > 0, 0 < x < 1 = ln(x) < 0, ln(1/x) = ln(x), ln (x) = 1/x > 0 so ln(x) is strictly increasing and ln (x) = 1/x 2 < 0, so the graph of ln(x) is concave downward. Use the Fundamental Theorem of Calculus and a change of variables. Example 8. h(x) = ln(2x) = h (x) = 1/x. Theorem 9. If a > 0 and b > 0, then: (1) ln(ab) = ln(a) + ln(b); (2) ln(1/b) = ln(b); (3) ln(a/b) = ln(a) ln(b); (4) ln(a n ) = n ln(a) for every integer n; (5) ln(a 1/q ) = (1/q) ln(a) for all q > 0; (6) ln(a r ) = r ln(a) for all r Q. (1). Let h(x) = ln(ax) = h (x) = (ln) (ax)(ax) = 1 a = 1/x = ax ln (x) = ln(ax) = h(x) = ln(x) + C = ln(a) = ln(a 1) = ln(1) + C = 0 + C = C = ln(ax) = ln(x) + ln(a) = ln(ab) = ln(a) + ln(b). (2) 0 = ln(1) = ln((1/a)a) = ln(1/a) + ln(a) = ln(1/a) = ln(a). (3) ln(a/b) = ln(a(1/b)) = ln(a) + ln(1/b) = ln(a) ln(b). (4) ln(a 1 ) = ln(a) = 1 ln(a) and inductively ln(a n+1 ) = ln(a n a) = ln(a n ) + ln(a) = n ln(a) + ln(a) = (n + 1) ln(a) for every n 1. ln(a 0 ) = ln(1) = 0 = 0 ln(a) and ln(a n ) = ln(1/a n ) = ln(a n ) = n ln(a) = ( n) ln(a). (5) ln(a) = ln((a 1/q ) q ) = q ln(a 1/q ) = ln(a 1/q ) = (1/q) ln(a). (6) If r = p/q, where p Z and q N, then ln(a r ) = ln((a 1/q ) p ) = p ln(a 1/q ) = p(1/q) ln(a) = (p/q) ln(a) = r ln(a). 9

11 Lemma 10. ln(x) + as x + and ln(x) as x 0+. ln(2) = 2 dw/w > 0 = 1 ln(2n ) = n ln(2) + as n +. Since ln(x), we have that ln(x) + as x +. ln(x) as x 0+ since ln(x) = ln(1/x). Theorem 11. ln : 0, + R is one-to-one and onto and therefore ln 1 exists. ln so ln is one-to-one. Since ln takes arbitrarily large positive and negative values and is continuous since it is differentiable, every horizontal line cuts the graph of ln. Therefore ln : 0, + R is onto. Theorem 12. f /f = ln( f ) + C and (1/u)du = ln( u ) + C Use the chain rule on ln(f) if f > 0 and on ln( f) if f < 0. Example 13. dx = ln(4/5) x x 2 5 Example 14. dθ = 2 ln(5). π/2 π/2 4 cos(θ) 3+2 sin(θ) Example 15. tan(x)dx = ln( sec(x) ) + C and cot(x)dx = ln( csc(x) ) + C. Example 16. sec(x)dx = ln( sec(x) + tan(x) ) + C and csc(x)dx = ln( csc(x) + cot(x) ) + C. For sec(x)dx multiply top and bottom by sec(x) + tan(x) and use f /f = ln( f ) + C. Similarly for csc(x)dx. Example 17. π/6 0 tan(2x)dx = ln(2)/2. 10

12 Example 18 (Logarithmic Differentiation). Find dy/dx if y = (x2 +1)(x+3) 1/2 x 1 for x > 1. Example 19. ln(x)dx = x ln(x) x + C Example 20. Let f(x) = (ln(x 3 + 6x 2 + 4)) 17. Compute f (x). Recall that ln : 0, + R is one-to-one and onto. Hence, ln 1 : R 0, + exists. Definition 21. exp(x) = ln 1 (x). Theorem 22. If x, y R, then: (1) exp(0) = 1; (2) exp(x + y) = exp(x) exp(y); (3) exp( x) = 1/ exp(x); (4) exp(rx) = (exp(x)) r for all r Q; (5) (exp) (x) = exp(x); (6) a r = exp(r ln(a)) for all a > 0 and r Q. (1) ln(exp(0)) = 0 = ln(1) = exp(0) = 1. (2) ln(exp(x + y)) = x + y and ln(exp(x) exp(y)) = ln(exp(x)) + ln(exp(y)) = x + y = exp(x + y) = exp(x) exp(y). (3) 1 = exp(0) = exp(x + ( x)) = exp(x) exp( x) = exp( x) = 1/ exp(x). (4) ln(exp(rx)) = rx and ln((exp(x)) r ) = r ln(exp(x)) = rx = exp(rx) = (exp(x)) r. (5) (exp) (x) = 1/(ln) (exp(x)) = 1/(1/ exp(x)) = exp(x). (6) ln(a r ) = r ln(a) and ln(exp(r ln(a))) = r ln(a) = a r = exp(r ln(a)). Corollary 23. If f is a real-valued function, then (exp(f)) = exp(f)f and exp(f)f = exp(f) + C. In particular, if a R, a 0, then (exp(ax)) = a exp(ax) and exp(ax)dx = exp(ax)/a + C. Use chain rule. 11

13 Exercise 2. Look over Section 6.2*. You should should be able to do #1 - #10, #15 - #44, #47 - #52, #65 - #74, #77 - #81, pp Graded: #9, #27, #31, #44, #50, #64, #70, pp These homeworks are due at the beginning of class on Wednesday, January 30, Remember that homeworks make up circa 40% of your grade. Exercise 3. Look over Section 6.3*. You should should be able to do #2 - #12, #15 - #16, #21 - #63, #67 - #70, #83 - #94, pp Graded: #26, #27, #41, #48, #59, #69, #93, pp These homeworks are due at the beginning of class on Wednesday, January 30, Remember that homeworks make up circa 40% of your grade. 12

14 Section 6.4*. General Logarithmic and Exponential Functions. Definition 24. e = exp(1) > 0 e = ln 1 (1) ln(e) = 1. Lemma 25. If r Q, then e r = exp(r) = ln 1 (r). e r = (exp(1)) r = exp(r 1) = exp(r). Heuristics. If a > 0 and b R, then y = a b ln(y) = ln(a b ) = b ln(a) a b = y = exp(ln(y)) = exp(b ln(a)). Definition 26. If a > 0 and b R, then a b = exp(b ln(a)). Proposition 27. ln(a b ) = b ln(a). ln(a b ) = ln(exp(b ln(a))) = b ln(a). Proposition 28. e x = exp(x). e x = exp(x ln(e)) = exp(x 1) = exp(x). Examples 29. Compute d dx e x and d dx esin(x). Heuristics. If a > 0 and b > 0, then we want a log a (b) = b = ln(a log a (b) ) = ln(b) = log a (b) ln(a) = ln(b) = log a (b) = ln(b)/ ln(a). Definition 30. If a > 0, a 1, and b > 0, let log a (b) = ln(b)/ ln(a). 13

15 Theorem 31. Let a > 0, a 1, b R, x > 0, and y > 0. Then: (1) log a (1) = 0; (2) log a (a) = 1; (3) log e (x) = ln(x); (4) log a (xy) = log a (x) + log a (y); (5) log a (1/x) = log a (x); (6) log a (x/y) = log a (x) log a (y); (7) log a (x b ) = b log a (x) = log a (a b ) = b; (8) (log a ) (x) = 1/x ln(a); (9) (d/dx)(log x (a)) = ln(a)/x ln 2 (x) (1) log a (1) = ln(1)/ ln(a) = 0/ ln(a) = 0. (2) log a (a) = ln(a)/ ln(a) = 1. (3) log e (x) = ln(x)/ ln(e) = ln(x)/1 = ln(x). (4) log a (xy) = ln(xy)/ ln(a) = ln(x)+ln(y) ln(a) = ln(x) ln(a) + ln(y) ln(a) = log a(x) + log a (y). (5) 0 = log a (1) = log a (x (1/x)) = log a (x) + log a (1/x) = log a (1/x) = log a (x). (6) log a (x/y) = log a (x(1/y)) = log a (x) + log a (1/y) = log a (x) log a (y). (7) log a (x b ) = ln(x b )/ ln(a) = b ln(x)/ ln(a) = b log a (x) = log a (a b ) = b log a (a) = b 1 = b. (8) (log a ) (x) = (ln(x)/ ln(a)) = 1/x ln(a). (9) (d/dx)(log x (a)) = (d/dx)(ln(a)/ ln(x)) = ln(a)/x ln 2 (x). Theorem 32. Let a > 0 and b, c, and x R. Then: (1) a log a (b) = b if a 1 and b > 0; (2) log a (a x ) = x if a 1; (3) a 0 = 1 and 1 x = 1; (4) a 1 = a; (5) a b a c = a b+c ; (6) (a b ) c = a bc ; (7) a b = 1/a b ; (8) a b /a c = a b c ; (9) if f(x) = a x, then f (x) = a x ln(a); (10) if f(x) = x b for x > 0, then f (x) = bx b 1 ; (1) a log a (b) = exp(log a (b) ln(a)) = exp((ln(b)/ ln(a)) ln(a)) = exp(ln(b)) = b. 14

16 (2) log a (a x ) = ln(a x )/ ln(a) = x ln(a)/ ln(a) = x. (3) a 0 = exp(0 ln(a)) = exp(0) = 1 and 1 x = exp(x ln(1)) = exp(x 0) = exp(0) = 1. (4) a 1 = exp(1 ln(a)) = exp(ln(a)) = a. (5) a b a c = exp(b ln(a)) exp(c ln(a)) = exp(b ln(a) + c ln(a)) = exp((b + c) ln(a)) = a b+c. (6) (a b ) c = exp(c ln(a b )) = exp(cb ln(a)) = exp((bc) ln(a)) = a bc. (7) 1 = a 0 = a b+( b) = a b a b = a b = 1/a b. (8) a b /a c = a b (1/a c ) = a b a c = a b+( c) = a b c. (9) f(x) = a x = exp(x ln(a)) = f (x) = (exp) (x ln(a))(x ln(a)) = a x ln(a). (10) f(x) = x b = exp(b ln(x)) = f (x) = (exp) (b ln(x))(b ln(x)) = x b b(1/x) = bx b x 1 = bx b 1. Theorem 33. lim x + (1 + 1 x )x = lim x 0+ (1 + x) 1/x = e and more generally lim x + (1 + R x )x = lim n + (1 + R n )n = e R for all R > 0. lim x + (1 + 1 x )x = lim x 0+ (1 + x) 1/x by making the subsitution x 1/x. exp is differentiable and therefore everywhere continuous. If f(x) = ln(1 + x) = f (x) = 1 f (0) = 1. exp(lim x 0 ln(1+x) Therefore 1 = f (0) = lim x 0 f(x) f(0) x 0 1+x = = lim x 0 ln(1+x) x = e = exp(1) = ) = exp(lim x x 0 ln(1 + x) 1/x ) = lim x 0 exp(ln(1 + x) 1/x ) = lim x 0 (1 + x) 1/x. Furthermore lim x + (1+ R x )x = lim x + ((1+ 1 (x/r) )(x/r) ) R = (lim x + (1+ 1 (x/r) )(x/r )) R = e R. Example 34. Let f(x) = x ln(x) for x > 0. Compute f (x). Show all steps. Example 35. Let f(x) = x x for x > 0. Compute f (x). Show all steps. Exercise 4. Look over Section 6.4*. You should should be able to do #3 - #10, #17 - #19, #21 - #42, #45 - #52, #54 - #62, pp Graded: #9, #30, #40, #50, #54, #56, pp These homeworks are due at the beginning of class on Wednesday, January 30, Remember that homeworks make up circa 40% of your grade. 15

17 Section 6.5. Exponential Growth and Decay. Suppose that we wish to solve y (t) = Ky(t) and y(0) = y 0. A simple calculation shows that (y(t)e Kt ) = y (t)e Kt + y(t)(e Kt ) = y (t)e Kt + y(t)( K)e Kt = Ky(t)e Kt + y(t)( K)e Kt = 0 = y(t)e Kt = C, a constant. Therefore y(t) = Ce Kt = y 0 = y(0) = Ce K0 = C = y(t) = y 0 e Kt. The Law of Exponential Change: y = y 0 e Kt, growth if K > 0 and decay if K < 0. Examples from book, population growth, continuously compounded interest, radioactivity, Newton s Law of cooling. Newton s Law of Cooling. Let T (t) be the temperature of a heated plate (ceramic or metal, for example) and toss it into a large pool (or ocean) of a fluid bath with constant temperature T B. We assume that the pool (or ocean) is so large that we can safely ignore a change in the bath s temperature from the hot plate. Then Newton s Law of Cooling states that T (t) = K(T (t) T B ), where K > 0 is a constant dependent on the materials used in the plate and the bath but independent for the range of temperatures (which can be very wide) under consideration. But then d dt (T (t) T B) = K(T (t) T B ) = T (t) T B = e Kt (T (0) T B ) = T (t) = T B + e Kt (T (0) T B ). Notice that as time goes by T (t) converges to T B but never quite gets there (assuming we know, as we will prove, that e Kt 0 as t +. Exercise 5. Look over Section 6.5. You should should be able to do #1 - #22, pp Graded: #3, #11, pp These homeworks are due at the beginning of class on Wednesday, February 6, Remember that homeworks make up circa 40% of your grade. 16

18 Section 6.6. Inverse Trigonometric Functions. Theorem 36. (1) cos : [0, π] [ 1, 1] is strictly decreasing and onto = arccos : [ 1, 1] [0, π]; (2) sin : [ π/2, π/2] [ 1, 1] is strictly increasing and onto = arcsin : [ 1, 1] [ π/2, π/2]; (3) tan : π/2, π/2 R is strictly increasing and onto = arctan : R π/2, π/2 ; (4) sec : [0, π] {π/2} { y y 1 } is one-to-one and onto = arcsec : { y y 1 } [0, π] {π/2}; (5) csc : [ π/2, π/2] {0} { y y 1 } is one-to-one and onto = arccsc : { y y 1 } [ π/2, π/2] {0}; (6) cot : 0, π R is strictly decreasing and onto = arccot : R 0, π. Theorem 37. (1) arccos(x) = π/2 arcsin(x); (2) arccot(x) = π/2 arctan(x); (3) arccsc(x) = π/2 arcsec(x) pseudo- Draw a picture of a right triangle and reason from there. Theorem 38. π = arctan(1)+arctan(2)+arctan(3), π/4 = arctan(1/2)+arctan(1/3) and π/4 = 4 arctan(1/5) arctan(1/239). We did this last semester. Theorem 39. (1) (arcsin) (x) = 1/ 1 x 2 ; (2) (arctan) (x) = 1/(1 + x 2 ); (3) (arcsec) (x) = 1/ x x

19 (arcsin) (x) = 1/ sin (arcsin(x)) = 1/ cos(arcsin(x)) = 1/ cos 2 (arcsin(x)) = 1/ 1 sin 2 (arcsin(x)) = 1/ 1 x 2 because cos is nonnegative on the range of arcsin. (arctan) (x) = 1/ tan (arctan(x)) = 1/ sec 2 (arctan(x)) = 1/(1 + tan 2 (arctan(x))) = 1/(1 + x 2 ). (arcsec) (x) = 1/ sec (arcsec(x)) = 1/ sec(arcsec(x)) tan(arcsec(x)) = 1/ sec(arcsec(x)) tan 2 (arcsec(x)) = 1/ x sec 2 (arcsec(x)) 1 = 1/ x x 2 1 since sec(w) > 0 and tan(w) > 0 on 0, π/2 and sec(w) < 0 and tan(w) < 0 on π/2, π. Corollary 40. (1) ((arcsin)(f)) = f / 1 f 2 = f / 1 f 2 = arcsin(f) + C; (2) ((arctan)(f)) = f /(1 + f 2 ) = f /(1 + f 2 ) = arctan(f) + C; (3) ((arcsec)(f)) = f / f f 2 1 = f / f f 2 1 = arcsec(f) + C Definitions and identities. arccos(x) + arccos( x) = π. arcsin(x) + arccos(x) = π/2. Examples on page 475. Theorem 41 (Gaussian Probability Density). + e x2 dx = π and therefore + exp( (x µ) 2 /2σ 2 ) 2πσ dx = 1 for µ R and σ > 0. 18

20 Sketch of Proof #1: The classical result that + exp( x2 )dx = π makes use of both Fubini s Theorem and the theorem on change of variables from Cartesian to polar coordinates, the latter of which (unlike the former) is rarely proved rigorously in a classroom setting. Therefore there perhaps is an interest in proofs, such as the following one, which avoid making use of such a two-dimensional change of variables. As usual let I = + exp( x2 )dx and note that Therefore I = π. I 2 /4 = ( = = = = = 1 2 = 1 2 = π [ [ [ [ exp( x 2 )dx)( exp( y 2 )dy) 0 ] exp( (x 2 + y 2 ))dy dx 0 ] exp( (x 2 (1 + (y/x) 2 ))dy dx 0 ] x exp( (x 2 (1 + u 2 ))du dx 0 ] x exp( (x 2 (1 + u 2 ))dx du [ exp( (x 2 (1 + u 2 )) du 1 + u 2 (1 + u 2 ) ] x=+ exp( (x µ) 2 /2σ 2 ) 2πσ dx = 1, giv- A simple change of variables in this integral proves that + ing the normal or Gaussian probability density. x=0 du Proof #2: Let I = + exp( x2 )dx and φ : R 0, + [0, + be the nonnegative infinitely differentiable rapidly decreasing function defined by φ(x, t) = exp( t(1 + x 2 )). φ is continuous. By Fubini s Theorem we have that + ( + φ(x, t)dt)dx = + ( + φ(x, t)dx)dt

21 But + ( + 0 φ(x, t)dt)dx = = = ( + 0 exp( t(1 + x 2 ))dt)dx ( exp( t(1 + x2 ) (1 + x 2 ) dx 1 + x 2 = arctan(x) + = π/2 ( π/2) = π and using simple changes of variables we have that + 0 Example 42. Find e x 1+e 2x dx. ( + φ(x, t)dx)dt = = = I = 2I = I 2. ( )dx exp( t(1 + x 2 ))dx)dt ( exp( t) + exp( x 2 )dx)dt t exp( t) dt t exp( t 2 )dt Exercise 6. Look over Section 6.6. You should should be able to do #1 - #14, #17 - #40, #43 - #50, #57 - #75, #77 - #80, pp Graded: #34, #40, #47, #49, #69, #74, #78, pp These homeworks are due at the beginning of class on Wednesday, February 13, Remember that homeworks make up circa 40% of your grade. 20

22 Section 6.8. Indeterminant Forms and L Hôpital s Rule We throughly covered these topics last semester but we can now encounter more sophisticated examples since we now know ln and exp. Go over some examples. Exercise 7. Look over Section 6.8. You should should be able to do #8 - #68, #73 - #76, #89, #93 - #103, pp Graded: #41, #42, #51, #53, #64, #93, #97, pp These homeworks are due at the beginning of class on Wednesday, February 13, Remember that homeworks make up circa 40% of your grade. 21

23 Lemma 43. exp(x) 1 + x for all x 0 = exp(x) + as x + and therefore exp(x) 0+ as x. Let f(x) = exp(x) x for x 0. f(0) = 1 and f (x) = exp(x) = 0 = f(x) 1 = exp(x) 1 + x. Lemma 44. α > 0 and x 1 = x α = exp(α ln(x)) 1 + α ln(x) = x α + as x +. Lemma 45. x 1 = 0 ln(x) x 1 < x. For x 1 we have that 0 ln(x) = x dw/w x dw = x 1 < x. This also follows 1 1 from the Mean Value Theorem. Fix 0 < δ < 1, for example δ = 1/2, in the next few elementary lemmas. Lemma 46. For x 1 we have that 0 ln(x)/x (1/δ)x (1 δ). For example, if δ = 1/2, then for x 1 we have that 0 ln(x)/x 2/ x. For x 1 we have that 0 ln(x) = (1/δ) ln(x δ ) (1/δ)x δ = 0 ln(x)/x (1/δ)x (1 δ). Lemma 47. For x 1 and α > 0 we have that 0 ln(x)/x α (1/αδ)x α(1 δ) ). For example, if δ = 1/2, then for x 1 and α > 0 we have that 0 ln(x)/x α (2/α) (1/x α/2 ). If α > 0 and x 1 we have that x α 1 = 0 ln(x α )/x α (1/δ)x α(1 δ). Lemma 48. For x 1, α > 0 and β > 0 we have that 0 ln β (x)/x α (β/αδ) β x α(1 δ). For example, if δ = 1/2, then for x 1, α > 0 and β > 0 we have that 0 ln β (x)/x α (2β/α) β (1/x α/2 ). 22

24 For x 1, α > 0 and β > 0 we have that 0 ln β (x)/x αβ (1/αδ) β x αβ(1 δ). Now replace α with α/β. Corollary 49. For x > 0, α > 0 and β > 0 we have that 0 x β exp( αx) (β/αδ) β exp( α(1 δ)x). For example, if δ = 1/2, then for x > 0, α > 0 and β > 0 we have that 0 x β exp( αx) (2β/α) β exp( αx/2). Replace x with exp(x). Corollary 50. For x > 0, α > 0, β > 0 and γ > 0 we have that 0 x β exp( αx γ ) (β/αδγ) β/γ exp( α(1 δ)x γ ). In particular, if δ = 1/2, then for x > 0, α > 0, β > 0 and γ > 0 we have that 0 x β exp( αx γ ) (2β/αγ) β/γ exp( αx γ /2). Replacing x with x γ we have that 0 x βγ exp( αx γ ) (β/αδ) β exp( α(1 δ)x γ ). Now replace β with β/γ. 23

25 Section 7.1. Integration by Parts. (fg) = f g+fg = f g = (fg) fg = f g = (fg) fg = f g = fg fg. Proposition 51. (1) ln(x)dx = x ln(x) x + C; (2) arctan(x)dx = x arctan(x) 1 2 ln(1 + x2 ) + C; (3) arcsin(x)dx = x arcsin(x) + 1 x 2 + C; (4) arcsec(x)dx = x arcsec(x) ln( x + x 2 1) + C. ln(x)dx = 1 ln(x)dx = x ln(x) x 1 x dx = x ln(x) dx = x ln(x) x + C. arctan(x)dx = 1 arctan(x)dx 1 = x arctan(x) x 1 + x dx 2 = x arctan(x) 1 2x x dx 2 = x arctan(x) 1 2 ln(1 + x2 ) + C. arcsin(x)dx = 1 arcsin(x)dx = 1 x arcsin(x) x 1 x 2 = x arcsin(x) + (1/2)(1 x 2 ) 1/2 ( 2x)dx = x arcsin(x) + 1 x 2 + C. 24

26 For x 1 we have arcsec(x)dx = 1 arcsec(x)dx 1 = x arcsec(x) x x x 2 1 dx 1 = x arcsec(x) x2 1 dx 1 = x arcsec(x) sec(θ) tan(θ)dθ sec(θ)2 1 1 = x arcsec(x) sec(θ) tan(θ)dθ tan(θ) = x arcsec(x) sec(θ)dθ = x arcsec(x) ln( sec(θ) + tan(θ) ) = x arcsec(x) ln( sec(θ) + sec 2 (θ) 1 ) = x arcsec(x) ln( x + x 2 1 ) = x arcsec(x) ln(x + x 2 1) = x arcsec(x) ln( x + x 2 1) + C 25

27 and for x 1 we have arcsec(x)dx = 1 arcsec(x)dx 1 = x arcsec(x) x x x 2 1 dx 1 = x arcsec(x) + x2 1 dx 1 = x arcsec(x) + sec(θ) tan(θ)dθ sec(θ)2 1 1 = x arcsec(x) + sec(θ) tan(θ)dθ tan(θ) = x arcsec(x) sec(θ)dθ Go over Examples #1 - #6. = x arcsec(x) ln( sec(θ) + tan(θ) ) = x arcsec(x) ln( sec(θ) sec 2 (θ) 1 ) = x arcsec(x) ln( x x 2 1 ) = x arcsec(x) ln( x + x 2 1) = x arcsec(x) ln( x + x 2 1) + C. Example 52. Find x 13 ln(x)dx. Give reasons for your answer. Exercise 8. Look over Section 7.1. You should should be able to do #1 - #58, #61 - #74, pp Graded: #21, #24, #30, #41, #42, #65, #67, pp These homeworks are due at the beginning of class on Wednesday, February 13, Remember that homeworks make up circa 40% of your grade. Reduction formula for sin, cos, sec and tan. Apply to Wallis Product Formula for π. 26

28 Proposition 53. x n e x dx = e x + C if n = 0 and x n e x dx = x n e x n x n 1 e x dx if n 1. Proposition 54 (Reduction Formula for tan). tan n (x)dx = x if n = 0, tan n (x)dx = ln( sec(x) ) + C if n = 1 and tan n (x)dx = tan n 1 (x) n 1 tan n 2 (x)dx + C if n 2 Proposition 55 (Reduction Formula for sec). Let n 3. Then sec n (x)dx = 1 n 1 secn 2 (x) tan(x) + n 2 n 1 sec n 2 (x)dx + C. Using integration by parts we deduce sec n (x)dx = sec n 2 (x) sec 2 (x)dx = sec n 2 (x)(1+ tan 2 (x))dx = sec n 2 (x)dx+ sec n 3 sec(x) tan(x) tan(x)dx = sec n 2 (x)dx+{ 1 1 n 2 n 1 n 2 n 2 secn 2 (x) tan(x) sec n 2 (x) sec 2 (x))dx} = sec n 2 (x)dx+ 1 n 2 secn 2 (x) tan(x) 1 n 2 sec n (x)dx. Hence, sec n (x)dx = 1 n 2 secn 2 (x) tan(x) + sec n 2 (x)dx and therefore sec n (x)dx = 1 n 1 secn 2 (x) tan(x) + n 2 sec n 2 (x)dx + C n 1 Corollary 56. sec 3 (x)dx = 1 sec(x) tan(x) + 1 ln( sec(x) + tan(x) ) + C 2 2 Proposition 57 (Reduction Formula for sin). Let n 2. Then sin n (x)dx = 1 n sinn 1 (x) cos(x) + n 1 n sin n 2 (x)dx + C. Using integration by parts we deduce sin n (x)dx = sin n 2 (x) sin 2 (x)dx = sin n 2 (x)(1 cos 2 (x))dx = sin n 2 (x)dx sin n 2 cos(x) cos(x)dx = sin n 2 (x)dx { 1 n 1 sinn 1 (x) cos(x) 1 n 1 sin n 1 (x)( sin(x))dx} = sin n 2 (x)dx 1 n 1 sinn 1 (x) cos(x) 1 n 1 sin n (x)dx. Hence, n n 1 sin n (x)dx = sin n 2 (x)dx 1 n 1 sinn 1 (x) cos(x) and therefore sin n (x)dx = 1 n sinn 1 (x) cos(x) + n 1 sin n 2 (x)dx. n Proposition 58 (Reduction Formula for cos). Let n 2. Then cos n (x)dx = 1 n cosn 1 (x) sin(x) + n 1 n cos n 2 (x)dx + C. 27

29 The proof is very similar to the reduction formula for sin. Theorem 59 (Wallis Product Formula). π = lim 4l 2 2 n + 1 l n = lim 4l 2 1 n n 2n (2n 1)(2n+1) For n 0 define I n = π/2 sin n (x)dx. Therefore I 0 0 = π and I 2 1 = 1. Use Proposition 57 to prove that I n+1 = n I n+1 n 1 for n 1. sin n+1 (x) sin n (x) sin n 1 (x) for all x [0, π/2]. Hence, 0 < I n+1 < I n < I n 1 and therefore 1 < I n /I n+1 < for n 1. Use induction to n prove that I 2k+1 = 2k 2k 2 2k and I 2k+1 2k 1 2k k = 2k 1 2k 3 2k π for each k 0. 2k 2k 2 2k Take the quotient of these two quantities to obtain π I2k+1 2 I 2k = k 2k and (2k 1)(2k+1) therefore conclude that π = lim 4l 2 2 n + 1 l n. 4l 2 1 Corollary 60. π = limn + (n!) 2 2 2n (2n)! n. Multiply top and bottom of Wallis Product Formula by n 2n. Lemma 61. n N = log(1 + 1 n ) > 2 2n+1. Let f(x) = log(1+ 1) 2 for x > 0. Compute that (Df)(x) = x 2x+1 1/x(x+1)(2x+1)2 < 0 for x > 0 and that lim x + f(x) = 0 = f(x) > 0 for x > 0. This lemma has a simple intepretation relating the area under the graph of 1 on [n, n+1] with the area of an inscribed x trapezoid on [n, n + 1]. Lemma 62. Let a n = n! (n/e) n n. Then a n and therefore lim n + a n exists. a n a n+1 = ( n )n+ 2 e > 1 by Lemma 61. Hence, a n. Lemma 63. x > 1 = log(x) x+ 1 2 log(t)dt. 2 x

30 This lemma has an obvious intepretation by comparing the area under the graph of log(t) on [x 1, x + 1 ] with the area of the circumscribing trapezoid. If x > 0, define 2 2 φ(t) = t +log(x) 1 log(t) for t > 0. Calculate that (Dφ)(t) = 1 1 and x x t (D2 φ)(t) = 1 > 0. t 2 Since (Dφ)(t) = 0 if and only if t = x and φ(x) = 0, we have that φ(t) 0 = log(x) x+ 1 2 x 1 2 [log(t) + 1 t x ]dt = x+ 1 2 x 1 2 Lemma 64. a n 1 for n N. n log(t)dt. log(x)dx = n D(x log(x) x)dx = n log(n) n + 1 = 1 1 log((n/e)n ) + 1. n log(x)dx = log(x)dx+ 1 2 l n 1 log(x)dx+ n log(x)dx n l n 1 log(l)+ 1 log(n) = 2 2 l+ 1 2 l log(n!) log( n). Hence log((n/e) n ) log(n!/ n) = (n/e) n n/n! 1 = a n 1. Theorem 65 (Stirling s Formula - James Stirling, 1764). lim n + n! (n/e) n 2πn = 1. Let L = lim n + a n 1. a 2 n a 2n 1 2 = (n!)2 2 2n (2n)! n L/ 2 = L 2 /L 2 on the other by Corollary 60. Hence, L = 2π. Corollary , 551, 179, 875, 952 = ( 52 25) converges to both π on one hand and to 29

31 Problem 1. Compute lim x 0+ (sin(x)) x. Problem 2. Compute Sample Questions for Test #1 on Friday, February 22, e x 1+e 2x dx. Problem 3. Compute the derivative of f(x) = tan(x) sec(x) for 0 < x < π/2. Problem 4. Compute dx. x 2 +2x+5 Problem 5. Compute arctan(x)dx. Problem 6. Use integration by parts to compute sin(x)e x dx. Problem 7. Use integration by parts to calculate arcsin(x)dx. Problem 8. Compute the derivative of f(x) = (sec(x)) tan(x) for 0 < x < π/2. Problem 9. Compute lim x 0+ x x. Problem 10. Compute e 2x +e 2x e 2x e 2x dx. Problem 11. Compute the derivative of f(x) = (ln(x)) x2 for x > 1. Problem 12. Compute 6dx 25 x 2. Problem 13. Compute x sec 2 (x)dx. 30

32 Problem 14. Calculate the limit lim x 0 (sin(x)/x) 1/x2. Problem 15. Compute lim x 0+ (sin(x)) sin(x). Problem 16. Compute e x +e x e x e x dx. Problem 17. Compute the derivative of f(x) = (tan(x)) sin(x) for 0 < x < π/2. Problem 18. Compute dx 8 6x x 2. Problem 19. Compute the derivative of f(x) = sin(x) cos(x) for 0 < x < π/2. Problem 20. Compute dx. x 2 +6x+10 Problem 21. Use integration by parts to compute cos(x)e x dx. Problem 22. Use logarithmic differentiation to compute the derivative of y = 3 x(x+1)(x 2) (x 2 +1)(2x+3) Problem 23. Compute dx. 1+e x Problem 24. Compute dx 4x x 2. Problem 25. Compute dx x. x Problem 26. Use integration by parts to deduce a reduction formula for ln n (x)dx for n 1. for x > 2. 31

33 Problem 27. Calculate 5 2 dx. 9+(x 2) 2 Problem 28. Use integration by parts to calculate x2 x dx. Problem 29. Compute xdx 1 x 4. Problem 30. Compute the derivative of f(x) = arctan(x) tan(x) for 0 < x < π/2. Problem 31. Compute dx. 4x 2 +4x+2 Problem 32. Compute 1 csc(x)+cot(x) dx. Problem 33. Use integration by parts to calculate x cos(x)dx. Problem 34. Use logarithmic differentiation to compute the derivative of f(x) = (x3 1) 4 3x 1) x Problem 35. Compute lim x + (1 + 1 x )x. Problem 36. Compute dx. x 2 +2x+2 Problem 37. Let f(x) = ln( x4 ). Specify the domain of f. On what intervals does f increase? decrease? x 1 Find the extreme values of f. Determine the concavity of the graph and find the points of inflection. Specify the asymptotes, if any. Problem 38. Calculate 1 x(1+ x) dx. Problem 39. Calculate x 1+x x dx. 32

34 Problem 40. Calculate 1 x ln 2 (x) dx. Problem 41. Compute e x x dx. Problem 42. Compute f (x) if f(x) = ln(cos(e 2x )). Problem 43. Compute e 1/x dx. x 2 Problem 44. Compute e x dx. e x +1 Problem 45. Compute (x x )dx. Problem 46. Compute 5p x+1 x+1 dx. Problem 47. Compute f (x) if f(x) = c 2 x 2 + c arcsin(x/c) if c > 0. Problem 48. Compute f (x) if f(x) = Problem 49. Compute Problem 50. Compute Problem 51. Compute dx (x 3). x 2 6x+8 dx x 1 ln. 2 (x) dx. x[1+ln 2 (x)] x c arcsin(x/c) if c > 0. 2 x 2 Problem 52. Show that arctan(x) + arccot(x) = π/2 for all x R. 33

35 Problem 53. Show that arcsec(x) + arccsc(x) = π/2 for x 1. Problem 54. Compute lim x 0 arcsin(x) x. Problem 55. Compute sec( x) x dx. Problem 56. Compute 1 x 2 4x+13 dx. Problem 57. Compute x x 2 dx. Problem 58. Given a < 1 find the value of b for which 1 0 Problem 59. Compute x dx. x 2 +2x+5 Problem 60. Compute cos( x)dx. Problem 61. Compute sin( x)dx. Problem 62. Compute x 2 e x dx. Problem 63. Calculate x 5 cos(x 3 )dx. Problem 64. Calculate cos(ln(x))dx. Problem 65. Calculate sin(ln(x))dx. Problem 66. Compute ln( x + 1)dx. 34 bdx 1 b 2 x 2 = a 0 dx 1 x 2.

36 Problem 67. Compute lim x 0 e x 1 x x 2. Problem 68. Compute lim x 0 e x2 1+x 2 x 4. Problem 69. Compute the derivative of f(x) = x sin(x) ( (x2 +4)(x π +1) 2x+1 ) 1/2. Problem 70. Compute the derivative of f(x) = x1/x e x (x 3 1) (x 2 +1) 1/2. Problem 71. Compute lim x + ln(1+1/x) sin(1/x). Problem 72. Compute lim x + [ln(x + 1) ln(x 1)]. Problem 73. Compute lim x 0 ( tan(x) x ) 1/x2. Problem 74. Compute lim x 1+ (ln(x)) sin(x 1). Problem 75. Compute e x dx. e 2x +2e x 3 Problem 76. Compute x sin(ax)dx if a 0. Problem 77. Compute x 1/2 e x1/2 dx. Problem 78. Find a reduction formula for x m (ln(x)) n dx if m and n are positive integers. Problem 79. Find a reduction formula for cos n (x)dx if n 2 is a positive integer. Problem 80. Compute π/2 0 cos 2n+1 (x)dx. 35

37 Problem 81. Suppose that f is a one-to-one functions with f(2) = 8 and f (2) = 4. What is the value of (f 1 ) (8)? Problem 82. Compute lim x 0 2 x 1 x. Problem 83. Compute d dx (sin(sin(ex ))). Problem 84. Compute ln 2 (x)+2 ln(x) 1 dx. x 36

38 Section 7.2. Trigonometric Integrals. Go over Examples #1 - #9. You should should be able to do Exercise 9. Look over Section 7.2 and the asssociated exercises. You should should be able to do #9 - #50, #55 - #58, #61 - #70, on pp Graded: #27, #47, #58, #63, #69, #70, pp These homeworks are due at the beginning of class on Wednesday, March 6, Remember that homeworks make up circa 40% of your grade. 37

39 Section 7.3. Trigonometric Substitutions. Go over examples #1 - #7. Exercise 10. Look over Section 7.3 and the asssociated exercises. You should should be able to do #1 - #44, pp Graded: #13, #25, #31(a), #32(a), #35, #40, pp These homeworks are due at the beginning of class on Wednesday, March 6, Remember that homeworks make up circa 40% of your grade. 38

40 Synthetic Division Let p(x) = 0 l n p lx l be an n-th degree polynomial with complex coefficients. We wish to test if λ is a root of p(x), i.e., we wish to test if p(λ) = 0. We will do even more, viz., we will find an efficient algorithm to calculate an (n 1)-st degree polynomial q(x) and a constant R so that p(x) = q(x)(x λ) + R. Then λ is a root of p(x) if and only if R = 0. Let q(x) = 0 l n 1 q lx l and substitute into the desired relation p(x) = q(x)(x λ) + R = (x λ) 0 l n 1 q lx l + R = 0 l n 1 q lx l+1 0 l n 1 λq lx l + R = 1 l n q l 1x l 0 l n 1 λq lx l + R = q n 1 x n + 1 l n 1 (q l 1 λq l )x l + (R λq 0 ). Equating coefficients of various powers of x in this equation we obtain: p n = q n 1 p n 1 = q n 2 λq n 1... p l = q l 1 λq l... p 1 = q 0 λq 1 p 0 = R λq 0 Rearranging terms we have: q n 1 = p n q n 2 = p n 1 + λq n 1 q l 1 = p l + λq l 39

41 q 0 = p 1 + λq 1 R = p 0 + λq 0 This recursive calculation for the coefficients of q(x) is most efficient when put into a tableau format. 40

42 Section 7.4. Partial Fractions. Go over Examples #1 - #9. More examples: (1) dx ; x 2 4 (2) dx ; x 3 x (3) dx ; x 3 x 2 +x 1 (4) dx x 3 1 ; (5) 2x 2 +3 dx; x(x 1) 2 (6) 2x dx; (x+1)(x 2 +1) (7) dx ; x(x 2 +x+1) (8) 3x 4 +x 3 +20x 2 +3x+31dx (x+1)(x 2 +4) 2 Example 67. Find 1 dx. Give reasons for your answer. x 2 9x+20 Exercise 11. Look over Section 7.4 and the asssociated exercises. You should should be able to do #1 - #51, #53, #54, #57 - #68, #71 - #75, pp Graded: #22, #46, #58, #71, #75, pp These homeworks are due at the beginning of class on Wednesday, March 20, Remember that homeworks make up circa 40% of your grade. Go over the substitution z = tan(x/2), cos(x) = 1 z2, sin(x) = 2z and dx = 2dz, and 1+z 2 1+z 2 1+z 2 examples of dx and of dx 1+cos(x). 1+sin(x)+cos(x) z = tan(x/2) = dz = sec2 (x/2)dx = (1+tan2 (x/2))dx = (1+z2 )dx = dx = 2dz z 2 cos(x) = 2 cos 2 2 (x/2) 1 = 1 = 2 1 = 1 z2. sec 2 (x/2) 1+z 2 1+z 2 sin(x) = 2 sin(x/2) cos(x/2) = 2 tan(x/2) sec 2 (x/2) = 2z. 1+z 2 41

43 Section 7.8. Improper Integrals. Definition of improper integrals with infinite integration limits. Go over Examples #1 - #10, pp Integrands with infinite discontinuities. Tests for Convergence and Divergence. Example 68. Evaluate the following integral or state that it diverges: + 1 dx (1+x). x Exercise 12. Look over Section 7.8 and the asssociated exercises. You should should be able to do #1 - #3, #5 - #40, #49 - #71, #72 - #82, pp Graded: #26, #51, #58, #70, #73 These homeworks are due at the beginning of class on Wednesday, March 20, Remember that homeworks make up circa 40% of your grade. Lemma 69. If A B < + and f : [A, + R is continuous, then + f(x)dx exists if and only if A f(x)dx exists. + B Theorem 70 (Comparison Test). If A B < +, f, g : [A, + R are continuous and 0 f(x) g(x) for all x B, then + f(x)dx exists if + g(x)dx exists and + g(x)dx = + if + f(x)dx = +. A A A A Corollary 71 (Limit Comparison Test). If A B < +, f, g : [A, + R are continuous, g(x) > 0 for all x B and lim x + f(x)/g(x) = L > 0, then + f(x)dx exists if and only if + g(x)dx exists. A A Furthermore + f(x)dx = + exists if and only if + g(x)dx = + exists. A A Practical examples of Fresnel Integrals and Airy Integrals and functions. 42

44 Section 9.3. Separable Differential Equations. Separable equations. dy dx = f(x)g(y) dy/g(y) = f(x)dx dy/g(y) = f(x)dx + C Why does this dubious reasoning actually make sense? Let G(y) = dy/g(y), let F (x) = f(x)dx and let C be a constant. Then assuming the relation G(y) = F (x) + C implicitly defines y as a smooth function of x, we have G(y) = F (x) + C d dx (G(y)) = d (F (x) + C) dx G (y)y = F (x) y /g(y) = f(x) y = f(x)g(y) Examples: y = e 3x 5y or y = arctan(x)(y 2 + 1). The constant C can only be determined from an initial value. Go over examples #1 - #6 in Section 9.3. Exercise 13. Graded - remembrance of things past: #15, #19, p #11, #13, p Look over Section 9.3 and the asssociated exercises. You should should be able to do #1 - #22, #29 - #32 (no graphing), #33 - #35, #36 - #54, pp Graded: #9, #12, #15, #22, #34, #54 on pp These homeworks are due at the beginning of class on Wednesday, March 27, Remember that homeworks make up circa 40% of your grade. 43

45 Section 9.5. Linear Equations. Problem: find the general solution to the first order linear constant coefficient differential equation y = ky + b. If k = 0 then y = bx + C. If k 0 then dy dx = ky + b dy dx + ( k)y = b exp( ( k)dx = kx ( k)dx) = exp( kx) exp( kx) dy + ( k) exp( kx)y = exp( kx)b dx d dy (exp( kx)y) = dx dx ( b k exp( kx) exp( kx)y = b k exp( kx) + C y = b k + C exp(kx) Thus y = y(x) is a solution to y = ky + b for k 0 if and only y = b + C exp(kx) for some k choice of C. C can only be determined by specifying an initial value for y, for example, by specifying y(0) or y(1). dy First order linear equations. + P (x)y = Q(x). If µ(x) = exp( P (x)dx), then dx d (µy) = µ(x)q(x) and solve modulo integrations. Go over RLC circuits and modification of Ohm s dx Law. Example 72. Solve the differential equation y + 2y = e x. Example 73. Solve the differential equation xy + 2y = xe x, y(1) = 1. Go over examples #1 - #5 in Section

46 Exercise 14. Look over Section 9.5 and the asssociated exercises. You should should be able to do #1 - #20, #23 - #26, #33 - #37, pp Graded: #14, #19, #25, #33 on pp These homeworks are due at the beginning of class on Wednesday, March 27, Remember that homeworks make up circa 40% of your grade. 45

47 Section Approximating Functions with Polynomials Theorem 74 (Taylor s Formula with Remainder). Let n Z, n 0 and let f be (n + 1)-times continuously differentiable on the closed interval with endpoints a and x. Then f(x) = 0 l n (Dl f)(a)(x a) l /l! + (1/n!) x (x a t) n (D n+1 f)(t)dt. (Note that we do not assume that a < x or x < a, though of course one of these must hold if a x.) By the Fundamental Theorem of Calculus we have that f(x) f(a) = x a (Df)(t)dt or that f(x) = f(a) + x a (Df)(t)dt. Therefore Taylor s Formula with Remainder holds when n = 0. Now use integration by parts to obtain that x a (Df)(t)dt = (x t)(df)(t) x a + x a (x t)(d2 f)(t)dt = (x a)(df)(a) + x a (x t)(d2 f)(t)dt, f(x) = f(a) + (x a)(df)(a) + x a (x t)(d2 f)(t)dt, and therefore Taylor s Formula with Remainder holds for n = 1. Recursively, if Taylor s Formula with Remainder holds at stage n, then use integration by parts to obtain that (1/n!) x (x a t)n (D n+1 f)(t)dt = (1/(n + 1)!)(x t) n+1 (D n+1 f)(t) x a + (1/(n + 1)!) x (x a t) n+1 (D (n+1)+1 f)(t)dt = (x a) n+1 (D n+1 f)(a)/(n+1)!+(1/(n+1)!) x a (x t)n+1 (D (n+1)+1 f)(t)dt and therefore f(x) = 0 l n (Dl f)(a)(x a) l /l! +(x a) n+1 (D n+1 f)(a)/(n + 1)! + (1/(n + 1)!) x a (x t)n+1 (D (n+1)+1 f)(t)dt 46

48 = 0 l n+1 (Dl f)(a)(x a) l /l! + (1/(n + 1)!) x a (x t)n+1 (D (n+1)+1 f)(t)dt. A simple change of variables allows one to express the error term in Taylor s Formula in a different and on occasion somewhat more useful form. Let t = a + w(x a) = dt = (x a)dw. Therefore (1/n!) x (x a t)n (D n+1 f)(t)dt = (1/n!) 1 (x 0 a)n (1 w) n (D n+1 f)(a + w(x a))(x a)dw = ((x a) n+1 /n!) 1 0 (Dn+1 f)(a + w(x a))(1 w) n dw. Lemma 75. Let {z n } n 1 be a sequence of nonzero real (or even complex) numbers and suppose that there exist some fixed 0 λ < 1 and some N 1 such that z n+1 /z n λ for all n N. (Notice that this holds if z n+1 z n 0 as n +.) Then z n 0 as n +. z n = z 1 z 2 z 1 z 3 z 2 z N z N 1 z N+1 z N zn z n 1 z 1 z 2 z 1 z 3 z 2 z N z N 1 λ N λ n = Cλ n 0 as n +, where C > 0 is a constant independent of n N. 47