Calculus of Variations
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1 Approximating Integrals Using Taylor Polynomials 1 1.1 Definitions
Seunghee Ye Ma 8: Week 7 Nov 10 Week 7 Summary This week, we will learn how we can approximate integrals using Taylor series and numerical methods. Topics Page 1 Approximating Integrals using Taylor Polynomials 1 1.1 Definitions . .1 1.2 Examples . .2 1.3 Approximating Integrals . .3 2 Numerical Integration 5 1 Approximating Integrals using Taylor Polynomials 1.1 Definitions When we first defined the derivative, recall that it was supposed to be the \instantaneous rate of change" of a function f(x) at a given point c. In other words, f 0 gives us a linear approximation of f(x) near c: for small values of " 2 R, we have f(c + ") ≈ f(c) + "f 0(c) But if f(x) has higher order derivatives, why stop with a linear approximation? Taylor series take this idea of linear approximation and extends it to higher order derivatives, giving us a better approximation of f(x) near c. Definition (Taylor Polynomial and Taylor Series) Let f(x) be a Cn function i.e. f is n-times continuously differentiable. Then, the n-th order Taylor polynomial of f(x) about c is: n X f (k)(c) T (f)(x) = (x − c)k n k! k=0 The n-th order remainder of f(x) is: Rn(f)(x) = f(x) − Tn(f)(x) If f(x) is C1, then the Taylor series of f(x) about c is: 1 X f (k)(c) T (f)(x) = (x − c)k 1 k! k=0 Note that the first order Taylor polynomial of f(x) is precisely the linear approximation we wrote down in the beginning. -
The Divergence As the Rate of Change in Area Or Volume
The divergence as the rate of change in area or volume Here we give a very brief sketch of the following fact, which should be familiar to you from multivariable calculus: If a region D of the plane moves with velocity given by the vector field V (x; y) = (f(x; y); g(x; y)); then the instantaneous rate of change in the area of D is the double integral of the divergence of V over D: ZZ ZZ rate of change in area = r · V dA = fx + gy dA: D D (An analogous result is true in three dimensions, with volume replacing area.) To see why this should be so, imagine that we have cut D up into a lot of little pieces Pi, each of which is rectangular with width Wi and height Hi, so that the area of Pi is Ai ≡ HiWi. If we follow the vector field for a short time ∆t, each piece Pi should be “roughly rectangular”. Its width would be changed by the the amount of horizontal stretching that is induced by the vector field, which is difference in the horizontal displacements of its two sides. Thisin turn is approximated by the product of three terms: the rate of change in the horizontal @f component of velocity per unit of displacement ( @x ), the horizontal distance across the box (Wi), and the time step ∆t. Thus ! @f ∆W ≈ W ∆t: i @x i See the figure below; the partial derivative measures how quickly f changes as you move horizontally, and Wi is the horizontal distance across along Pi, so the product fxWi measures the difference in the horizontal components of V at the two ends of Wi. -
Multivariable Calculus Workbook Developed By: Jerry Morris, Sonoma State University
Multivariable Calculus Workbook Developed by: Jerry Morris, Sonoma State University A Companion to Multivariable Calculus McCallum, Hughes-Hallett, et. al. c Wiley, 2007 Note to Students: (Please Read) This workbook contains examples and exercises that will be referred to regularly during class. Please purchase or print out the rest of the workbook before our next class and bring it to class with you every day. 1. To Purchase the Workbook. Go to the Sonoma State University campus bookstore, where the workbook is available for purchase. The copying charge will probably be between $10.00 and $20.00. 2. To Print Out the Workbook. Go to the Canvas page for our course and click on the link \Math 261 Workbook", which will open the file containing the workbook as a .pdf file. BE FOREWARNED THAT THERE ARE LOTS OF PICTURES AND MATH FONTS IN THE WORKBOOK, SO SOME PRINTERS MAY NOT ACCURATELY PRINT PORTIONS OF THE WORKBOOK. If you do choose to try to print it, please leave yourself enough time to purchase the workbook before our next class in case your printing attempt is unsuccessful. 2 Sonoma State University Table of Contents Course Introduction and Expectations .............................................................3 Preliminary Review Problems ........................................................................4 Chapter 12 { Functions of Several Variables Section 12.1-12.3 { Functions, Graphs, and Contours . 5 Section 12.4 { Linear Functions (Planes) . 11 Section 12.5 { Functions of Three Variables . 14 Chapter 13 { Vectors Section 13.1 & 13.2 { Vectors . 18 Section 13.3 & 13.4 { Dot and Cross Products . 21 Chapter 14 { Derivatives of Multivariable Functions Section 14.1 & 14.2 { Partial Derivatives . -
A Quotient Rule Integration by Parts Formula Jennifer Switkes ([email protected]), California State Polytechnic Univer- Sity, Pomona, CA 91768
A Quotient Rule Integration by Parts Formula Jennifer Switkes ([email protected]), California State Polytechnic Univer- sity, Pomona, CA 91768 In a recent calculus course, I introduced the technique of Integration by Parts as an integration rule corresponding to the Product Rule for differentiation. I showed my students the standard derivation of the Integration by Parts formula as presented in [1]: By the Product Rule, if f (x) and g(x) are differentiable functions, then d f (x)g(x) = f (x)g(x) + g(x) f (x). dx Integrating on both sides of this equation, f (x)g(x) + g(x) f (x) dx = f (x)g(x), which may be rearranged to obtain f (x)g(x) dx = f (x)g(x) − g(x) f (x) dx. Letting U = f (x) and V = g(x) and observing that dU = f (x) dx and dV = g(x) dx, we obtain the familiar Integration by Parts formula UdV= UV − VdU. (1) My student Victor asked if we could do a similar thing with the Quotient Rule. While the other students thought this was a crazy idea, I was intrigued. Below, I derive a Quotient Rule Integration by Parts formula, apply the resulting integration formula to an example, and discuss reasons why this formula does not appear in calculus texts. By the Quotient Rule, if f (x) and g(x) are differentiable functions, then ( ) ( ) ( ) − ( ) ( ) d f x = g x f x f x g x . dx g(x) [g(x)]2 Integrating both sides of this equation, we get f (x) g(x) f (x) − f (x)g(x) = dx. -
Operations on Power Series Related to Taylor Series
Operations on Power Series Related to Taylor Series In this problem, we perform elementary operations on Taylor series – term by term differen tiation and integration – to obtain new examples of power series for which we know their sum. Suppose that a function f has a power series representation of the form: 1 2 X n f(x) = a0 + a1(x − c) + a2(x − c) + · · · = an(x − c) n=0 convergent on the interval (c − R; c + R) for some R. The results we use in this example are: • (Differentiation) Given f as above, f 0(x) has a power series expansion obtained by by differ entiating each term in the expansion of f(x): 1 0 X n−1 f (x) = a1 + a2(x − c) + 2a3(x − c) + · · · = nan(x − c) n=1 • (Integration) Given f as above, R f(x) dx has a power series expansion obtained by by inte grating each term in the expansion of f(x): 1 Z a1 a2 X an f(x) dx = C + a (x − c) + (x − c)2 + (x − c)3 + · · · = C + (x − c)n+1 0 2 3 n + 1 n=0 for some constant C depending on the choice of antiderivative of f. Questions: 1. Find a power series representation for the function f(x) = arctan(5x): (Note: arctan x is the inverse function to tan x.) 2. Use power series to approximate Z 1 2 sin(x ) dx 0 (Note: sin(x2) is a function whose antiderivative is not an elementary function.) Solution: 1 For question (1), we know that arctan x has a simple derivative: , which then has a power 1 + x2 1 2 series representation similar to that of , where we subsitute −x for x. -
Multivariable and Vector Calculus
Multivariable and Vector Calculus Lecture Notes for MATH 0200 (Spring 2015) Frederick Tsz-Ho Fong Department of Mathematics Brown University Contents 1 Three-Dimensional Space ....................................5 1.1 Rectangular Coordinates in R3 5 1.2 Dot Product7 1.3 Cross Product9 1.4 Lines and Planes 11 1.5 Parametric Curves 13 2 Partial Differentiations ....................................... 19 2.1 Functions of Several Variables 19 2.2 Partial Derivatives 22 2.3 Chain Rule 26 2.4 Directional Derivatives 30 2.5 Tangent Planes 34 2.6 Local Extrema 36 2.7 Lagrange’s Multiplier 41 2.8 Optimizations 46 3 Multiple Integrations ........................................ 49 3.1 Double Integrals in Rectangular Coordinates 49 3.2 Fubini’s Theorem for General Regions 53 3.3 Double Integrals in Polar Coordinates 57 3.4 Triple Integrals in Rectangular Coordinates 62 3.5 Triple Integrals in Cylindrical Coordinates 67 3.6 Triple Integrals in Spherical Coordinates 70 4 Vector Calculus ............................................ 75 4.1 Vector Fields on R2 and R3 75 4.2 Line Integrals of Vector Fields 83 4.3 Conservative Vector Fields 88 4.4 Green’s Theorem 98 4.5 Parametric Surfaces 105 4.6 Stokes’ Theorem 120 4.7 Divergence Theorem 127 5 Topics in Physics and Engineering .......................... 133 5.1 Coulomb’s Law 133 5.2 Introduction to Maxwell’s Equations 137 5.3 Heat Diffusion 141 5.4 Dirac Delta Functions 144 1 — Three-Dimensional Space 1.1 Rectangular Coordinates in R3 Throughout the course, we will use an ordered triple (x, y, z) to represent a point in the three dimensional space. -
Derivative of Power Series and Complex Exponential
LECTURE 4: DERIVATIVE OF POWER SERIES AND COMPLEX EXPONENTIAL The reason of dealing with power series is that they provide examples of analytic functions. P1 n Theorem 1. If n=0 anz has radius of convergence R > 0; then the function P1 n F (z) = n=0 anz is di®erentiable on S = fz 2 C : jzj < Rg; and the derivative is P1 n¡1 f(z) = n=0 nanz : Proof. (¤) We will show that j F (z+h)¡F (z) ¡ f(z)j ! 0 as h ! 0 (in C), whenever h ¡ ¢ n Pn n k n¡k jzj < R: Using the binomial theorem (z + h) = k=0 k h z we get F (z + h) ¡ F (z) X1 (z + h)n ¡ zn ¡ hnzn¡1 ¡ f(z) = a h n h n=0 µ ¶ X1 a Xn n = n ( hkzn¡k) h k n=0 k=2 µ ¶ X1 Xn n = a h( hk¡2zn¡k) n k n=0 k=2 µ ¶ X1 Xn¡2 n = a h( hjzn¡2¡j) (by putting j = k ¡ 2): n j + 2 n=0 j=0 ¡ n ¢ ¡n¡2¢ By using the easily veri¯able fact that j+2 · n(n ¡ 1) j ; we obtain µ ¶ F (z + h) ¡ F (z) X1 Xn¡2 n ¡ 2 j ¡ f(z)j · jhj n(n ¡ 1)ja j( jhjjjzjn¡2¡j) h n j n=0 j=0 X1 n¡2 = jhj n(n ¡ 1)janj(jzj + jhj) : n=0 P1 n¡2 We already know that the series n=0 n(n ¡ 1)janjjzj converges for jzj < R: Now, for jzj < R and h ! 0 we have jzj + jhj < R eventually. -
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 -
20-Finding Taylor Coefficients
Taylor Coefficients Learning goal: Let’s generalize the process of finding the coefficients of a Taylor polynomial. Let’s find a general formula for the coefficients of a Taylor polynomial. (Students complete worksheet series03 (Taylor coefficients).) 2 What we have discovered is that if we have a polynomial C0 + C1x + C2x + ! that we are trying to match values and derivatives to a function, then when we differentiate it a bunch of times, we 0 get Cnn!x + Cn+1(n+1)!x + !. Plugging in x = 0 and everything except Cnn! goes away. So to (n) match the nth derivative, we must have Cn = f (0)/n!. If we want to match the values and derivatives at some point other than zero, we will use the 2 polynomial C0 + C1(x – a) + C2(x – a) + !. Then when we differentiate, we get Cnn! + Cn+1(n+1)!(x – a) + !. Now, plugging in x = a makes everything go away, and we get (n) Cn = f (a)/n!. So we can generate a polynomial of any degree (well, as long as the function has enough derivatives!) that will match the function and its derivatives to that degree at a particular point a. We can also extend to create a power series called the Taylor series (or Maclaurin series if a is specifically 0). We have natural questions: • For what x does the series converge? • What does the series converge to? Is it, hopefully, f(x)? • If the series converges to the function, we can use parts of the series—Taylor polynomials—to approximate the function, at least nearby a. -
MULTIVARIABLE CALCULUS Sample Midterm Problems October 1, 2009 INSTRUCTOR: Anar Akhmedov
MULTIVARIABLE CALCULUS Sample Midterm Problems October 1, 2009 INSTRUCTOR: Anar Akhmedov 1. Let P (1, 0, 3), Q(0, 2, 4) and R(4, 1, 6) be points. − − − (a) Find the equation of the plane through the points P , Q and R. (b) Find the area of the triangle with vertices P , Q and R. Solution: The vector P~ Q P~ R = < 1, 2, 1 > < 3, 1, 9 > = < 17, 6, 5 > is the normal vector of this plane,× so equation− of− the− plane× is 17(x 1)+6(y −0)+5(z + 3) = 0, which simplifies to 17x 6y 5z = 32. − − − − − Area = 1 P~ Q P~ R = 1 < 1, 2, 1 > < 3, 1, 9 > = √350 2 | × | 2 | − − − × | 2 2. Let f(x, y)=(x y)3 +2xy + x2 y. Find the linear approximation L(x, y) near the point (1, 2). − − 2 2 2 2 Solution: fx = 3x 6xy +3y +2y +2x and fy = 3x +6xy 3y +2x 1, so − − − − fx(1, 2) = 9 and fy(1, 2) = 2. Then the linear approximation of f at (1, 2) is given by − L(x, y)= f(1, 2)+ fx(1, 2)(x 1) + fy(1, 2)(y 2)=2+9(x 1)+( 2)(y 2). − − − − − 3. Find the distance between the parallel planes x +2y z = 1 and 3x +6y 3z = 3. − − − Use the following formula to find the distance between the given parallel planes ax0+by0+cz0+d D = | 2 2 2 | . Use a point from the second plane (for example (1, 0, 0)) as (x ,y , z ) √a +b +c 0 0 0 and the coefficents from the first plane a = 1, b = 2, c = 1, and d = 1. -
The Set of Continuous Functions with Everywhere Convergent Fourier Series
TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 302, Number 1, July 1987 THE SET OF CONTINUOUS FUNCTIONS WITH EVERYWHERE CONVERGENT FOURIER SERIES M. AJTAI AND A. S. KECHRIS ABSTRACT. This paper deals with the descriptive set theoretic properties of the class EC of continuous functions with everywhere convergent Fourier se- ries. It is shown that this set is a complete coanalytic set in C(T). A natural coanalytic rank function on EC is studied that assigns to each / € EC a count- able ordinal number, which measures the "complexity" of the convergence of the Fourier series of /. It is shown that there exist functions in EC (in fact even differentiable ones) which have arbitrarily large countable rank, so that this provides a proper hierarchy on EC with uii distinct levels. Let G(T) be the Banach space of continuous 27r-periodic functions on the reals with the sup norm. We study in this paper descriptive set theoretic aspects of the subset EC of G(T) consisting of the continuous functions with everywhere convergent Fourier series. It is easy to verify that EC is a coanalytic set. We show in §3 that EC is actually a complete coanalytic set and therefore in particular is not Borel. This answers a question in [Ku] and provides another example of a coanalytic non-Borel set. We also discuss in §2 (see in addition the Appendix) the relations of this kind of result to the studies in the literature concerning the classification of the set of points of divergence of the Fourier series of a continuous function. -
Calculus of Variations Raju K George, IIST
Calculus of Variations Raju K George, IIST Lecture-1 In Calculus of Variations, we will study maximum and minimum of a certain class of functions. We first recall some maxima/minima results from the classical calculus. Maxima and Minima Let X and Y be two arbitrary sets and f : X Y be a well-defined function having domain → X and range Y . The function values f(x) become comparable if Y is IR or a subset of IR. Thus, optimization problem is valid for real valued functions. Let f : X IR be a real valued → function having X as its domain. Now x X is said to be maximum point for the function f if 0 ∈ f(x ) f(x) x X. The value f(x ) is called the maximum value of f. Similarly, x X is 0 ≥ ∀ ∈ 0 0 ∈ said to be a minimum point for the function f if f(x ) f(x) x X and in this case f(x ) is 0 ≤ ∀ ∈ 0 the minimum value of f. Sufficient condition for having maximum and minimum: Theorem (Weierstrass Theorem) Let S IR and f : S IR be a well defined function. Then f will have a maximum/minimum ⊆ → under the following sufficient conditions. 1. f : S IR is a continuous function. → 2. S IR is a bound and closed (compact) subset of IR. ⊂ Note that the above conditions are just sufficient conditions but not necessary. Example 1: Let f : [ 1, 1] IR defined by − → 1 x = 0 f(x)= − x x = 0 | | 6 1 −1 +1 −1 Obviously f(x) is not continuous at x = 0.