Math 2300: Calculus II Project 3: Comparison of Improper Integrals
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MATH 305 Complex Analysis, Spring 2016 Using Residues to Evaluate Improper Integrals Worksheet for Sections 78 and 79
MATH 305 Complex Analysis, Spring 2016 Using Residues to Evaluate Improper Integrals Worksheet for Sections 78 and 79 One of the interesting applications of Cauchy's Residue Theorem is to find exact values of real improper integrals. The idea is to integrate a complex rational function around a closed contour C that can be arbitrarily large. As the size of the contour becomes infinite, the piece in the complex plane (typically an arc of a circle) contributes 0 to the integral, while the part remaining covers the entire real axis (e.g., an improper integral from −∞ to 1). An Example Let us use residues to derive the formula p Z 1 x2 2 π 4 dx = : (1) 0 x + 1 4 Note the somewhat surprising appearance of π for the value of this integral. z2 First, let f(z) = and let C = L + C be the contour that consists of the line segment L z4 + 1 R R R on the real axis from −R to R, followed by the semi-circle CR of radius R traversed CCW (see figure below). Note that C is a positively oriented, simple, closed contour. We will assume that R > 1. Next, notice that f(z) has two singular points (simple poles) inside C. Call them z0 and z1, as shown in the figure. By Cauchy's Residue Theorem. we have I f(z) dz = 2πi Res f(z) + Res f(z) C z=z0 z=z1 On the other hand, we can parametrize the line segment LR by z = x; −R ≤ x ≤ R, so that I Z R x2 Z z2 f(z) dz = 4 dx + 4 dz; C −R x + 1 CR z + 1 since C = LR + CR. -
Section 8.8: Improper Integrals
Section 8.8: Improper Integrals One of the main applications of integrals is to compute the areas under curves, as you know. A geometric question. But there are some geometric questions which we do not yet know how to do by calculus, even though they appear to have the same form. Consider the curve y = 1=x2. We can ask, what is the area of the region under the curve and right of the line x = 1? We have no reason to believe this area is finite, but let's ask. Now no integral will compute this{we have to integrate over a bounded interval. Nonetheless, we don't want to throw up our hands. So note that b 2 b Z (1=x )dx = ( 1=x) 1 = 1 1=b: 1 − j − In other words, as b gets larger and larger, the area under the curve and above [1; b] gets larger and larger; but note that it gets closer and closer to 1. Thus, our intuition tells us that the area of the region we're interested in is exactly 1. More formally: lim 1 1=b = 1: b − !1 We can rewrite that as b 2 lim Z (1=x )dx: b !1 1 Indeed, in general, if we want to compute the area under y = f(x) and right of the line x = a, we are computing b lim Z f(x)dx: b !1 a ASK: Does this limit always exist? Give some situations where it does not exist. They'll give something that blows up. -
Math 142 – Quiz 6 – Solutions 1. (A) by Direct Calculation, Lim 1+3N 2
Math 142 – Quiz 6 – Solutions 1. (a) By direct calculation, 1 + 3n 1 + 3 0 + 3 3 lim = lim n = = − . n→∞ n→∞ 2 2 − 5n n − 5 0 − 5 5 3 So it is convergent with limit − 5 . (b) By using f(x), where an = f(n), x2 2x 2 lim = lim = lim = 0. x→∞ ex x→∞ ex x→∞ ex (Obtained using L’Hˆopital’s Rule twice). Thus the sequence is convergent with limit 0. (c) By comparison, n − 1 n + cos(n) n − 1 ≤ ≤ 1 and lim = 1 n + 1 n + 1 n→∞ n + 1 n n+cos(n) o so by the Squeeze Theorem, the sequence n+1 converges to 1. 2. (a) By the Ratio Test (or as a geometric series), 32n+2 32n 3232n 7n 9 L = lim (16 )/(16 ) = lim = . n→∞ 7n+1 7n n→∞ 32n 7 7n 7 The ratio is bigger than 1, so the series is divergent. Can also show using the Divergence Test since limn→∞ an = ∞. (b) By the integral test Z ∞ Z t 1 1 t dx = lim dx = lim ln(ln x)|2 = lim ln(ln t) − ln(ln 2) = ∞. 2 x ln x t→∞ 2 x ln x t→∞ t→∞ So the integral diverges and thus by the Integral Test, the series also diverges. (c) By comparison, (n3 − 3n + 1)/(n5 + 2n3) n5 − 3n3 + n2 1 − 3n−2 + n−3 lim = lim = lim = 1. n→∞ 1/n2 n→∞ n5 + 2n3 n→∞ 1 + 2n−2 Since P 1/n2 converges (p-series, p = 2 > 1), by the Limit Comparison Test, the series converges. -
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. -
1 Improper Integrals
July 14, 2019 MAT136 { Week 6 Justin Ko 1 Improper Integrals In this section, we will introduce the notion of integrals over intervals of infinite length or integrals of functions with an infinite discontinuity. These definite integrals are called improper integrals, and are understood as the limits of the integrals we introduced in Week 1. Definition 1. We define two types of improper integrals: 1. Infinite Region: If f is continuous on [a; 1) or (−∞; b], the integral over an infinite domain is defined as the respective limit of integrals over finite intervals, Z 1 Z t Z b Z b f(x) dx = lim f(x) dx; f(x) dx = lim f(x) dx: a t!1 a −∞ t→−∞ t R 1 R a If both a f(x) dx < 1 and −∞ f(x) dx < 1, then Z 1 Z a Z 1 f(x) dx = f(x) dx + f(x) dx: −∞ −∞ a R 1 If one of the limits do not exist or is infinite, then −∞ f(x) dx diverges. 2. Infinite Discontinuity: If f is continuous on [a; b) or (a; b], the improper integral for a discon- tinuous function is defined as the respective limit of integrals over finite intervals, Z b Z t Z b Z b f(x) dx = lim f(x) dx f(x) dx = lim f(x) dx: a t!b− a a t!a+ t R c R b If f has a discontinuity at c 2 (a; b) and both a f(x) dx < 1 and c f(x) dx < 1, then Z b Z c Z b f(x) dx = f(x) dx + f(x) dx: a a c R b If one of the limits do not exist or is infinite, then a f(x) dx diverges. -
Series: Convergence and Divergence Comparison Tests
Series: Convergence and Divergence Here is a compilation of what we have done so far (up to the end of October) in terms of convergence and divergence. • Series that we know about: P∞ n Geometric Series: A geometric series is a series of the form n=0 ar . The series converges if |r| < 1 and 1 a1 diverges otherwise . If |r| < 1, the sum of the entire series is 1−r where a is the first term of the series and r is the common ratio. P∞ 1 2 p-Series Test: The series n=1 np converges if p1 and diverges otherwise . P∞ • Nth Term Test for Divergence: If limn→∞ an 6= 0, then the series n=1 an diverges. Note: If limn→∞ an = 0 we know nothing. It is possible that the series converges but it is possible that the series diverges. Comparison Tests: P∞ • Direct Comparison Test: If a series n=1 an has all positive terms, and all of its terms are eventually bigger than those in a series that is known to be divergent, then it is also divergent. The reverse is also true–if all the terms are eventually smaller than those of some convergent series, then the series is convergent. P P P That is, if an, bn and cn are all series with positive terms and an ≤ bn ≤ cn for all n sufficiently large, then P P if cn converges, then bn does as well P P if an diverges, then bn does as well. (This is a good test to use with rational functions. -
Math 1B, Lecture 11: Improper Integrals I
Math 1B, lecture 11: Improper integrals I Nathan Pflueger 30 September 2011 1 Introduction An improper integral is an integral that is unbounded in one of two ways: either the domain of integration is infinite, or the function approaches infinity in the domain (or both). Conceptually, they are no different from ordinary integrals (and indeed, for many students including myself, it is not at all clear at first why they have a special name and are treated as a separate topic). Like any other integral, an improper integral computes the area underneath a curve. The difference is foundational: if integrals are defined in terms of Riemann sums which divide the domain of integration into n equal pieces, what are we to make of an integral with an infinite domain of integration? The purpose of these next two lecture will be to settle on the appropriate foundation for speaking of improper integrals. Today we consider integrals with infinite domains of integration; next time we will consider functions which approach infinity in the domain of integration. An important observation is that not all improper integrals can be considered to have meaningful values. These will be said to diverge, while the improper integrals with coherent values will be said to converge. Much of our work will consist in understanding which improper integrals converge or diverge. Indeed, in many cases for this course, we will only ask the question: \does this improper integral converge or diverge?" The reading for today is the first part of Gottlieb x29:4 (up to the top of page 907). -
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 -
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 . -
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. -
Math 113 Lecture #16 §7.8: Improper Integrals
Math 113 Lecture #16 x7.8: Improper Integrals Improper Integrals over Infinite Intervals. How do we make sense of and evaluate an improper integral like Z 1 1 2 dx? 1 x Can we apply the Fundamental Theorem of Calculus to evaluate this? No, because the FTC only applies to integrals on finite intervals! So instead we convert the integral over an infinite interval into a limit of integrals over finite intervals: Z A 1 lim 2 dx: A!1 1 x Now we can apply the FTC to get Z A 1 1 A 1 lim dx = lim − = lim − + 1 = 1: A!1 2 A!1 A!1 1 x x 1 A Since this limit exists, we say that the improper integral converges, and the value of this limit we take as the value of the improper integral: Z 1 1 2 dx = 1: 1 x Z t Definition. If f(x) dx exists for every t ≥ a, then the improper integral a Z 1 Z A f(x) dx = lim f(x) dx a A!1 a converges provided the limit exists (as a finite number); otherwise it diverges. Z b If f(x) dx exists for every t ≤ b, then the improper integral t Z b Z b f(x) dx = lim f(x) dx 1 B!1 B converges provided the limit exists (as a finite number); otherwise it diverges. Z 1 Z a If both improper integrals f(x) dx and f(x) dx are convergent, then we define a −∞ Z 1 Z 1 Z a f(x) dx = f(x) dx + f(x) dx: −∞ a −∞ The choice of a is arbitrary. -
MAT137 Basic Comparison Test and Limit Comparison Test Watch Videos
MAT137 Course website http://uoft.me/MAT137 . Basic comparison test and limit comparison test Watch Videos 13.2–13.4 before next Monday’s class. Supplementary video: 13.1 Watch Videos 13.5–13.9 before next Wednesday’s class. Francisco Guevara Parra () MAT137 27 February 2019 1 / 11 A simple BCT application ∞ 1 dx We want to determine whether x Z1 x + e is convergent or divergent. We can try at least two comparisons: 1 1 1. Compare and . x x + ex 1 1 2. Compare and . ex x + ex Try both. What can you conclude from each one of them? Francisco Guevara Parra () MAT137 27 February 2019 2 / 11 True or False Let a R. ∈ Let f and g be continuous functions on [a, ). ∞ Assume that x a, 0 f (x) g(x) . ∀ ≥ ≤ ≤ What can we conclude? ∞ ∞ 1. IF f (x)dx is convergent, THEN g(x)dx is convergent. Za Za ∞ ∞ 2. IF f (x)dx = , THEN g(x)dx = . Za ∞ Za ∞ ∞ ∞ 3. IF g(x)dx is convergent, THEN f (x)dx is convergent. Za Za ∞ ∞ 4. IF g(x)dx = , THEN f (x)dx = . Za ∞ Za ∞ Francisco Guevara Parra () MAT137 27 February 2019 3 / 11 True or False - Part II Let a R. ∈ Let f and g be continuous functions on [a, ). ∞ Assume that x a, f (x) g(x) . ∀ ≥ ≤ What can we conclude? ∞ ∞ 1. IF f (x)dx is convergent, THEN g(x)dx is convergent. Za Za ∞ ∞ 2. IF f (x)dx = , THEN g(x)dx = . Za ∞ Za ∞ ∞ ∞ 3. IF g(x)dx is convergent, THEN f (x)dx is convergent.