Final Exam: Cheat Sheet"

Total Page:16

File Type:pdf, Size:1020Kb

Final Exam: Cheat Sheet 7/17/97 Final Exam: \Cheat Sheet" Z Z 1 2 6. csc xdx = cot x dx =lnjxj 1. x Z Z x a x 7. sec x tan xdx = sec x 2. a dx = ln a Z Z 8. csc x cot xdx = csc x 3. sin xdx = cos x Z Z 1 x dx 1 = tan 9. 4. cos xdx = sin x 2 2 x + a a a Z Z dx x 1 2 p = sin 10. 5. sec xdx = tan x 2 2 a a x Z Z d b [f y g y ] dy [f x g x] dx. In terms of y : A = 11. Area in terms of x: A = c a Z Z d b 2 2 [g y ] dy [f x] dx. Around the y -axis: V = 12. Volume around the x-axis: V = c a Z b 13. Cylindrical Shells: V = 2xf x dx where 0 a< b a Z Z b b b 0 0 14. f xg x dx = f xg x] f xg x dx a a a R m 2 2 n 2 2 15. HowtoEvaluate sin x cos xdx.:Ifm or n are o dd use sin x =1 cos x and cos x =1 sin x, 2 1 1 2 resp ectively and then apply substitution. Otherwise, use sin x = 1 cos 2x or cos x = 1 + cos 2x 2 2 1 or sin x cos x = sin 2x. 2 R m n 2 2 2 16. HowtoEvaluate tan x sec xdx.:Ifm is o dd or n is even, use tan x = sec x 1 and sec x = 2 1 + tan x, resp ectively and then apply substitution. 17. Table of Trigonometric Substitutions Expression Substitution Identity p 2 2 2 2 a x x = a sin , 1 sin = cos 2 2 p 2 2 2 2 < < a + x x = a tan , 1 + tan = sec 2 2 p 3 2 2 2 2 x a x = a sec 0 < or < sec 1 = tan 2 2 " n n1 X X x f x and T = f x +f x f x +2 18. M =x i n i n 0 n 2 i=1 i=1 x 19. S = [f x +4f x +2f x +4f x + +2f x + 4f x + f x ] n 0 1 2 3 n2 n1 n 3 3 3 5 K b a K b a K b a 2 2 4 20. jE j and jE j and jE j T M S 2 2 4 12n 24n 180n Z Z Z Z 1 t b b 21. f x dx = lim f x dx and f x dx = lim f x dx t!1 t!1 a a 1 t P 1 22. Test for Divergence. If lim a do es not exist or if lim a 6= 0, then the series a is n!1 n n!1 n n n=1 divergent. 23. Remainder Estimate for the Integral Test. If a converges by the Integral Test and R = s s , n n n Z Z 1 1 then f x dx R f x dx n n+1 n 10 P P 24. Limit Comparison Test. Supp ose that a and b are series with p ositive terms. a If n n a a n n = c>0, then either b oth series converge or b oth diverge. b If lim = 0 and lim n!1 n!1 b b P n P n P P a n b converges, then a also converges. c If lim = 1 and b diverges, then a also n n n!1 n n b n diverges. P n1 25. Alternating Series Estimation Theorem. If s = 1 b is the sum of an alternating series n that satis es a and b ab ove, then jR j = js s jb n n n+1 P 1 a n+1 a is absolutely .IfL exists and L<1, then the series 26. The Ratio Test Let L = lim n n!1 n=1 a n P 1 a is divergent. convergent. If L exists and L> 1 or is in nity, then the series n n=1 p P 1 n 27. The Ro ot Test Let L = lim j = L.IfL exists and L<1, then the series a is ja n!1 n n n=1 P 1 absolutely convergent. If L exists and L>1 or is in nity, then the series a is divergent. n n=1 P 1 n 28. Theorem. If f x= c x a has radius of convergence R> 0 then f x is di erentiable on n n=0 Z 1 1 n+1 X X x a 0 n1 , each a R; a + R and a f x= nc x a and b f x dx = C + c n n n +1 n=1 n=0 with radius of convergence R. n 29. Theorem. If f has a p ower series representation at a, that is, if f x=c x a for jx aj <R, n n f a then its co ecients are of the form n! 30. Taylor's Formula. If f has n + 1 derivatives in an interval I that contains the number a, then for x in I there is a number z strictly b etween x and a such that the remainder term in the Taylor series n+1 f z n+1 can b e expressed as R x= x a n n + 1! 1 1 1 2n+1 n X X X x 1 x n x 1 n 31. = x ,1; 1 ,1; 1 ,[1; 1] e = tan x = 1 1 x n! 2n +1 n=0 n=0 n=0 1 1 2n+1 2n X X x x n n ,1; 1 ,1; 1 sin x = cos x = 1 1 2n + 1! 2n! n=0 n=0 1 X k k n 32. The Binomial Series If k is any real numb er and jxj < 1, then 1 + x = x where n n=0 k k 1k n+1 k k = =1. for n 1 and n 0 n! dy dy =dt dx 33. = if 6=0. dx dx=dt dt 0 34. The Arc Length Formula. If f is continuous on [a; b], then the length of the curve y = f x, Z b p 0 2 a x b,is L = 1+ [f x] dx. a 35. Theorem. If a curve C is describ ed by the parametric equations x = f t, y = g t, t , 0 0 where f and g are continuous on [ ; ] and C is traversed exactly once as t increases from to , s Z 2 2 dx dy then the length of C is L = + dt. dt dt s Z 2 2 dx dy 2y + dt 36. Surface Area. S = dt dt y 2 2 2 . 37. Polar co ordinate conversion: x = r cos , y = r sin , r = x + y , and tan = x s Z Z 2 b b 1 dr 2 2 38. For p olar co ordinates: A = r d and L = r + d 2 d a a 11.
Recommended publications
  • Sequences and Series Pg. 1
    Sequences and Series Exercise Find the first four terms of the sequence 3 1, 1, 2, 3, . [Answer: 2, 5, 8, 11] Example Let be the sequence defined by 2, 1, and 21 13 2 for 0,1,2,, find and . Solution: (k = 0) 21 13 2 2 2 3 (k = 1) 32 24 2 6 2 8 ⁄ Exercise Without using a calculator, find the first 4 terms of the recursively defined sequence • 5, 2 3 [Answer: 5, 7, 11, 19] • 4, 1 [Answer: 4, , , ] • 2, 3, [Answer: 2, 3, 5, 8] Example Evaluate ∑ 0.5 and round your answer to 5 decimal places. ! Solution: ∑ 0.5 ! 0.5 0.5 0.5 ! ! ! 0.5 0.5 ! ! 0.5 0.5 0.5 0.5 0.5 0.5 0.041666 0.003125 0.000186 0.000009 . Exercise Find and evaluate each of the following sums • ∑ [Answer: 225] • ∑ 1 5 [Answer: 521] pg. 1 Sequences and Series Arithmetic Sequences an = an‐1 + d, also an = a1 + (n – 1)d n S = ()a +a n 2 1 n Exercise • For the arithmetic sequence 4, 7, 10, 13, (a) find the 2011th term; [Answer: 6,034] (b) which term is 295? [Answer: 98th term] (c) find the sum of the first 750 terms. [Answer: 845,625] rd th • The 3 term of an arithmetic sequence is 8 and the 16 term is 47. Find and d and construct the sequence. [Answer: 2,3;the sequence is 2,5,8,11,] • Find the sum of the first 800 natural numbers. [Answer: 320,400] • Find the sum ∑ 4 5.
    [Show full text]
  • 3.3 Convergence Tests for Infinite Series
    3.3 Convergence Tests for Infinite Series 3.3.1 The integral test We may plot the sequence an in the Cartesian plane, with independent variable n and dependent variable a: n X The sum an can then be represented geometrically as the area of a collection of rectangles with n=1 height an and width 1. This geometric viewpoint suggests that we compare this sum to an integral. If an can be represented as a continuous function of n, for real numbers n, not just integers, and if the m X sequence an is decreasing, then an looks a bit like area under the curve a = a(n). n=1 In particular, m m+2 X Z m+1 X an > an dn > an n=1 n=1 n=2 For example, let us examine the first 10 terms of the harmonic series 10 X 1 1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + + : n 2 3 4 5 6 7 8 9 10 1 1 1 If we draw the curve y = x (or a = n ) we see that 10 11 10 X 1 Z 11 dx X 1 X 1 1 > > = − 1 + : n x n n 11 1 1 2 1 (See Figure 1, copied from Wikipedia) Z 11 dx Now = ln(11) − ln(1) = ln(11) so 1 x 10 X 1 1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + + > ln(11) n 2 3 4 5 6 7 8 9 10 1 and 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + < ln(11) + (1 − ): 2 3 4 5 6 7 8 9 10 11 Z dx So we may bound our series, above and below, with some version of the integral : x If we allow the sum to turn into an infinite series, we turn the integral into an improper integral.
    [Show full text]
  • The Binomial Series 16.3   
    The Binomial Series 16.3 Introduction In this Section we examine an important example of an infinite series, the binomial series: p(p − 1) p(p − 1)(p − 2) 1 + px + x2 + x3 + ··· 2! 3! We show that this series is only convergent if |x| < 1 and that in this case the series sums to the value (1 + x)p. As a special case of the binomial series we consider the situation when p is a positive integer n. In this case the infinite series reduces to a finite series and we obtain, by replacing x with b , the binomial theorem: a n(n − 1) (b + a)n = bn + nbn−1a + bn−2a2 + ··· + an. 2! Finally, we use the binomial series to obtain various polynomial expressions for (1 + x)p when x is ‘small’. ' • understand the factorial notation $ Prerequisites • have knowledge of the ratio test for convergence of infinite series. Before starting this Section you should ... • understand the use of inequalities '& • recognise and use the binomial series $% Learning Outcomes • state and use the binomial theorem On completion you should be able to ... • use the binomial series to obtain numerical approximations & % 26 HELM (2008): Workbook 16: Sequences and Series ® 1. The binomial series A very important infinite series which occurs often in applications and in algebra has the form: p(p − 1) p(p − 1)(p − 2) 1 + px + x2 + x3 + ··· 2! 3! in which p is a given number and x is a variable. By using the ratio test it can be shown that this series converges, irrespective of the value of p, as long as |x| < 1.
    [Show full text]
  • Polar Coordinates and Complex Numbers Infinite Series Vectors And
    Contents CHAPTER 9 Polar Coordinates and Complex Numbers 9.1 Polar Coordinates 348 9.2 Polar Equations and Graphs 351 9.3 Slope, Length, and Area for Polar Curves 356 9.4 Complex Numbers 360 CHAPTER 10 Infinite Series 10.1 The Geometric Series 10.2 Convergence Tests: Positive Series 10.3 Convergence Tests: All Series 10.4 The Taylor Series for ex, sin x, and cos x 10.5 Power Series CHAPTER 11 Vectors and Matrices 11.1 Vectors and Dot Products 11.2 Planes and Projections 11.3 Cross Products and Determinants 11.4 Matrices and Linear Equations 11.5 Linear Algebra in Three Dimensions CHAPTER 12 Motion along a Curve 12.1 The Position Vector 446 12.2 Plane Motion: Projectiles and Cycloids 453 12.3 Tangent Vector and Normal Vector 459 12.4 Polar Coordinates and Planetary Motion 464 CHAPTER 13 Partial Derivatives 13.1 Surfaces and Level Curves 472 13.2 Partial Derivatives 475 13.3 Tangent Planes and Linear Approximations 480 13.4 Directional Derivatives and Gradients 490 13.5 The Chain Rule 497 13.6 Maxima, Minima, and Saddle Points 504 13.7 Constraints and Lagrange Multipliers 514 CHAPTER Infinite Series Infinite series can be a pleasure (sometimes). They throw a beautiful light on sin x and cos x. They give famous numbers like n and e. Usually they produce totally unknown functions-which might be good. But on the painful side is the fact that an infinite series has infinitely many terms. It is not easy to know the sum of those terms.
    [Show full text]
  • Beyond Mere Convergence
    Beyond Mere Convergence James A. Sellers Department of Mathematics The Pennsylvania State University 107 Whitmore Laboratory University Park, PA 16802 [email protected] February 5, 2002 – REVISED Abstract In this article, I suggest that calculus instruction should include a wider variety of examples of convergent and divergent series than is usually demonstrated. In particular, a number of convergent series, P k3 such as 2k , are considered, and their exact values are found in a k≥1 straightforward manner. We explore and utilize a number of math- ematical topics, including manipulation of certain power series and recurrences. During my most recent spring break, I read William Dunham’s book Euler: The Master of Us All [3]. I was thoroughly intrigued by the material presented and am certainly glad I selected it as part of the week’s reading. Of special interest were Dunham’s comments on series manipulations and the power series identities developed by Euler and his contemporaries, for I had just completed teaching convergence and divergence of infinite series in my calculus class. In particular, Dunham [3, p. 47-48] presents Euler’s proof of the Basel Problem, a challenge from Jakob Bernoulli to determine the 1 P 1 exact value of the sum k2 . Euler was the first to solve this problem by k≥1 π2 proving that the sum equals 6 . I was reminded of my students’ interest in this result when I shared it with them just weeks before. I had already mentioned to them that exact values for relatively few families of convergent series could be determined.
    [Show full text]
  • Series Convergence Tests Math 121 Calculus II Spring 2015
    Series Convergence Tests Math 121 Calculus II Spring 2015 Some series converge, some diverge. Geometric series. We've already looked at these. We know when a geometric series 1 X converges and what it converges to. A geometric series arn converges when its ratio r lies n=0 a in the interval (−1; 1), and, when it does, it converges to the sum . 1 − r 1 X 1 The harmonic series. The standard harmonic series diverges to 1. Even though n n=1 1 1 its terms 1, 2 , 3 , . approach 0, the partial sums Sn approach infinity, so the series diverges. The main questions for a series. Question 1: given a series does it converge or diverge? Question 2: if it converges, what does it converge to? There are several tests that help with the first question and we'll look at those now. The term test. The only series that can converge are those whose terms approach 0. That 1 X is, if ak converges, then ak ! 0. k=1 Here's why. If the series converges, then the limit of the sequence of its partial sums n th X approaches the sum S, that is, Sn ! S where Sn is the n partial sum Sn = ak. Then k=1 lim an = lim (Sn − Sn−1) = lim Sn − lim Sn−1 = S − S = 0: n!1 n!1 n!1 n!1 The contrapositive of that statement gives a test which can tell us that some series diverge. Theorem 1 (The term test). If the terms of the series don't converge to 0, then the series diverges.
    [Show full text]
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • A Tentative Interpretation of the Epistemological Significance of the Encrypted Message Sent by Newton to Leibniz in October 1676
    Advances in Historical Studies 2014. Vol.3, No.1, 22-32 Published Online February 2014 in SciRes (http://www.scirp.org/journal/ahs) http://dx.doi.org/10.4236/ahs.2014.31004 A Tentative Interpretation of the Epistemological Significance of the Encrypted Message Sent by Newton to Leibniz in October 1676 Jean G. Dhombres Centre Alexandre Koyré-Ecole des Hautes Etudes en Sciences Sociales, Paris, France Email: [email protected] Received October 8th, 2013; revised November 9th, 2013; accepted November 16th, 2013 Copyright © 2014 Jean G. Dhombres. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In accordance of the Creative Commons Attribution License all Copyrights © 2014 are reserved for SCIRP and the owner of the intellectual property Jean G. Dhombres. All Copyright © 2014 are guarded by law and by SCIRP as a guardian. In Principia mathematica philosophiæ naturalis (1687), with regard to binomial formula and so general Calculus, Newton claimed that Leibniz proposed a similar procedure. As usual within Newtonian style, no explanation was provided on that, and nowadays it could only be decoded by Newton himself. In fact to the point, not even he gave up any mention of Leibniz in 1726 on the occasion of the third edition of the Principia. In this research, I analyse this controversy comparing the encrypted passages, and I will show that at least two ideas were proposed: vice versa and the algebraic handing of series. Keywords: Leibniz; Newton; Calculus Principia; Binomial Series; Controversy; Vice Versa Introduction tober 1676 to Leibniz.
    [Show full text]
  • Sequences, Series and Taylor Approximation (Ma2712b, MA2730)
    Sequences, Series and Taylor Approximation (MA2712b, MA2730) Level 2 Teaching Team Current curator: Simon Shaw November 20, 2015 Contents 0 Introduction, Overview 6 1 Taylor Polynomials 10 1.1 Lecture 1: Taylor Polynomials, Definition . .. 10 1.1.1 Reminder from Level 1 about Differentiable Functions . .. 11 1.1.2 Definition of Taylor Polynomials . 11 1.2 Lectures 2 and 3: Taylor Polynomials, Examples . ... 13 x 1.2.1 Example: Compute and plot Tnf for f(x) = e ............ 13 1.2.2 Example: Find the Maclaurin polynomials of f(x) = sin x ...... 14 2 1.2.3 Find the Maclaurin polynomial T11f for f(x) = sin(x ) ....... 15 1.2.4 QuestionsforChapter6: ErrorEstimates . 15 1.3 Lecture 4 and 5: Calculus of Taylor Polynomials . .. 17 1.3.1 GeneralResults............................... 17 1.4 Lecture 6: Various Applications of Taylor Polynomials . ... 22 1.4.1 RelativeExtrema .............................. 22 1.4.2 Limits .................................... 24 1.4.3 How to Calculate Complicated Taylor Polynomials? . 26 1.5 ExerciseSheet1................................... 29 1.5.1 ExerciseSheet1a .............................. 29 1.5.2 FeedbackforSheet1a ........................... 33 2 Real Sequences 40 2.1 Lecture 7: Definitions, Limit of a Sequence . ... 40 2.1.1 DefinitionofaSequence .......................... 40 2.1.2 LimitofaSequence............................. 41 2.1.3 Graphic Representations of Sequences . .. 43 2.2 Lecture 8: Algebra of Limits, Special Sequences . ..... 44 2.2.1 InfiniteLimits................................ 44 1 2.2.2 AlgebraofLimits.............................. 44 2.2.3 Some Standard Convergent Sequences . .. 46 2.3 Lecture 9: Bounded and Monotone Sequences . ..... 48 2.3.1 BoundedSequences............................. 48 2.3.2 Convergent Sequences and Closed Bounded Intervals . .... 48 2.4 Lecture10:MonotoneSequences .
    [Show full text]
  • Calculus Online Textbook Chapter 10
    Contents CHAPTER 9 Polar Coordinates and Complex Numbers 9.1 Polar Coordinates 348 9.2 Polar Equations and Graphs 351 9.3 Slope, Length, and Area for Polar Curves 356 9.4 Complex Numbers 360 CHAPTER 10 Infinite Series 10.1 The Geometric Series 10.2 Convergence Tests: Positive Series 10.3 Convergence Tests: All Series 10.4 The Taylor Series for ex, sin x, and cos x 10.5 Power Series CHAPTER 11 Vectors and Matrices 11.1 Vectors and Dot Products 11.2 Planes and Projections 11.3 Cross Products and Determinants 11.4 Matrices and Linear Equations 11.5 Linear Algebra in Three Dimensions CHAPTER 12 Motion along a Curve 12.1 The Position Vector 446 12.2 Plane Motion: Projectiles and Cycloids 453 12.3 Tangent Vector and Normal Vector 459 12.4 Polar Coordinates and Planetary Motion 464 CHAPTER 13 Partial Derivatives 13.1 Surfaces and Level Curves 472 13.2 Partial Derivatives 475 13.3 Tangent Planes and Linear Approximations 480 13.4 Directional Derivatives and Gradients 490 13.5 The Chain Rule 497 13.6 Maxima, Minima, and Saddle Points 504 13.7 Constraints and Lagrange Multipliers 514 CHAPTER Infinite Series Infinite series can be a pleasure (sometimes). They throw a beautiful light on sin x and cos x. They give famous numbers like n and e. Usually they produce totally unknown functions-which might be good. But on the painful side is the fact that an infinite series has infinitely many terms. It is not easy to know the sum of those terms.
    [Show full text]
  • Chapter 10 Infinite Series
    10.1 The Geometric Series (page 373) CHAPTER 10 INFINITE SERIES 10.1 The Geometric Series The advice in the text is: Learn the geometric series. This is the most important series and also the simplest. The pure geometric series starts with the constant term 1: 1+x +x2 +... = A.With other symbols CzoP = &.The ratio between terms is x (or r). Convergence requires 1x1 < 1(or Irl < 1). A closely related geometric series starts with any number a, instead of 1. Just multiply everything by a : a + ox + ax2 + - .= &. With other symbols x:=oern = &. As a particular case, choose a = xN+l. The geometric series has its first terms chopped off: Now do the opposite. Keep the first terms and chop of the last terms. Instead of an infinite series you have a finite series. The sum looks harder at first, but not after you see where it comes from: 1- xN+l N 1- rN+l Front end : 1+ x + - -.+ xN = or Ern= 1-2 - ' n=O 1-r The proof is in one line. Start with the whole series and subtract the tail end. That leaves the front end. It is =N+1 the difference between and TT; - complete sum minus tail end. Again you can multiply everything by a. The front end is a finite sum, and the restriction 1x1 < 1or Irl < 1 is now omitted. We only avoid x = 1or r = 1, because the formula gives g. (But with x = 1the front end is 1+ 1+ . + 1. We don't need a formula to add up 1's.) Personal note: I just heard a lecture on the mathematics of finance.
    [Show full text]
  • 1. the Root Test Video: Root Test Proof
    LECTURE 20: SERIES (II) Let's continue our series extravaganza! Today's goal is to prove the celebrated Ratio and Root Tests and to compare them. 1. The Root Test Video: Root Test Proof Among all the convergence tests, the root test is the best one, or at least better than the ratio test. Let me remind you how it works: Example 1: Use the root test to figure out if the following series converges: 1 X n 3n n=0 n Let an = 3n , then the root test tells you to look at: 1 1 1 1 n n n n n n n!1 1 janj n = = = ! = α < 1 n n 1 3 3 ( n ) 3 3 P Therefore an converges absolutely. 1 Since limn!1 janj n doesn't always exist, we need to replace this with 1 n lim supn!1 janj , which always exists. Therefore, we obtain the root Date: Wednesday, May 13, 2020. 1 2 LECTURE 20: SERIES (II) test: Root Test P Consider an and let 1 α = lim sup janj n n!1 P P (1) If α < 1, then an converges absolutely (that is janj converges) P (2) If α > 1, then an diverges (3) If α = 1, then the root test is inconclusive, meaning that you'd have to use another test Proof of (1): (α < 1 ) converges absolutely) 1 n Main Idea: Since lim supn!1 janj = α < 1, then for large n we will 1 n P P n have janj n ≤ α. So janj ≤ α and therefore janj ≤ α , which is a geometric series that converges, since α < 1.
    [Show full text]