...the students acquainted with the concept of basic topics from Mathematics, which they need to pursue their Engineering degree in different disciplines. Course Contents: Module I: Differential Calculus Successive differentiation, Leibnitz’s theorem (without proof), Mean value theorem, Taylor’s theorem (proof), Remainder terms, Asymptote & Curvature, Partial derivatives, Chain rule, Differentiation of Implicit functions, Exact differentials, Tangents and Normals, Maxima, Approximations, Differentiation under integral sign, Jacobians and transformations of coordinates. Module II: Integral Calculus Fundamental theorems, Reduction formulae, Properties of definite integrals, Applications to length, area, volume, surface of revolution, improper integrals, Multiple Integrals-Double integrals, Applications to areas, volumes. Module III: Ordinary Differential Equations Formation of ODEs, Definition of order, degree & solutions, ODE of first order : Method of separation of variables, homogeneous and non homogeneous equations, Exactness & integrating factors, Linear equations & Bernoulli equations, General linear ODE of nth order, Solution of homogeneous equations, Operator method, Method of undetermined coefficients, Solution of simple simultaneous ODE. Module IV: Vector Calculus Scalar and Vector Field, Derivative of a Vector, Gradient, Directional Derivative, Divergence and Curl and their Physical Significance...
Words: 310 - Pages: 2
...Math 5616H Midterm 1 with solutions Spring 2013 March 8, 2013 Total 80 points 1. (15 points) Let f (x) and g(x) be real continuous functions on an interval [a, b], such that b b f 2 (x) dx = a a b g 2 (x) dx = 1. Prove that a f (x)g(x) dx ≥ −1, and that a b f (x)g(x) dx = −1 if and only if f ≡ −g on [a, b]. Answer: Since f and g are continuous, so is (f + g)2 , which is therefore integrable. We compute: b b b b b 0≤ a b [f (x)+g(x)]2 dx = a f (x)2 dx+2 a f (x)g(x) dx+ a g(x)2 dx = 1+2 a f (x)g(x) dx+1, so a f (x)g(x) dx ≥ −1. If it is = −1, then the first “≤” must be “=”, so the continuous function [f (x) + g(x)]2 ≡ 0, and f ≡ −g on [a, b]. 2. (25 points) Let α(x) be a strictly increasing function on the interval [0, 1], such that α(0) = 0 and α(1) = 1. Show that the Riemann-Stieltjes integral 1 α(x) dα(x), 0 exists if and only if α is continuous on [0, 1], and evaluate this integral if it is continuous. Answer: Consider any partition P of [0, 1] : P = {0 = x0 , . . . , 1 = xn }. Since α is increasing, Mi := supx∈[xi−1 ,xi ] α(x) = α(xi ) and mi = α(xi−1 ). Then n n U (P, α, α) − L(P, α, α) = i=1 (Mi − mi )∆αi = i=1 (∆αi )2 . Suppose α is continuous; then since [0, 1] is compact, α is uniformly continuous. Thus, for any given ε > 0 there is δ > 0 so that if |x − y| < δ then |α(x) − α(y)| < ε. Hence if P ∗ is a refinement of P which satisfies xi − xi−1 < δ for all i = 1, . . . , n, we have n U (P ∗ ...
Words: 966 - Pages: 4
...A Taylor series for the function arctan The integral If we invert y = arctan(x) to obtain x = tan y, then, by differentiating with respect to y, we find dx/dy = sec2 y = 1 + tan2 y = 1 + x2 . Thus we have (ignoring the constant of integration) y = arctan(x) = dx . 1 + x2 (1) If we now differentiate y = arctan(x/a) with respect to x, where a is a constant, we have, by the chain rule, y = 1 a 1 a = 2 . 1 + (x/a)2 x + a2 (2) Thus we obtain the indefinite integral 1 dx = arctan(x/a). x 2 + a2 a (3) The Taylor Series By expanding the integrand in (3) as a geometric series 1/(1 − r) = 1 + r + r 2 + . . ., |r| < 1, and then integrating, we can obtain a series to represent the function arctan(x/a). We use the dummy variable t for the integration on [0, x] and we first write x x arctan(x/a) = a 0 dt 1 = t2 + a 2 a 0 dt 1 + (t/a)2 (4) Substituting the geometric series with r = −(t/a)2 , we find 1 arctan(x/a) = a x ∞ (−t2 /a2 )n dt = 0 n=0 (−1)n x 2n + 1 a n=0 ∞ 2n+1 . (5) The radius of convergence of this series is the same as that of the original geometric series, namely R = 1, or, in terms of x, |x/a| < 1. The series is a convergent alternating series at the right-hand end point x = a; and it can be shown that sum equals the value of arctan(1) = π/4 (as we might hope). Thus we have the nice (but slowly converging) series for π given by π 1 1 1 = 1 − + − + .... (6) 4 3 5 7 By choosing partial sums of (5) we obtain a sequence of Taylor polynomial...
Words: 350 - Pages: 2
...Mathematics Syllabus Algebra: Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations. Quadratic equations with real coefficients, relations between roots and coefficients, formation of quadratic equations with given roots, symmetric functions of roots. Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural numbers. Logarithms and their properties. Permutations and combinations, Binomial theorem for a positive integral index, properties of binomial coefficients. Matrices as a rectangular array of real numbers, equality of matrices, addition, multiplication by a scalar and product of matrices, transpose of a matrix, determinant of a square matrix of order up to three, inverse of a square matrix of order up to three, properties of these matrix operations, diagonal, symmetric and skew-symmetric matrices and their properties, solutions of simultaneous linear equations in two or three variables. Addition and multiplication rules of probability, conditional probability, Bayes Theorem, independence of events, computation of probability of events using permutations and combinations. Trigonometry: Trigonometric functions, their periodicity and graphs, addition...
Words: 631 - Pages: 3
...ESTABLISHING THE STRESS LOADING THE ELEMENTS [pic] Fig. 1.1 2. MAIN SCREW CALCULUS 2.1. CHOOSING THE MATERIAL It is chosen OL 50 STAS 500/2 [3] PRE-DIMENSIONING CALCULUS The calculus load F= Q·ctgαmin αmin= 30º [pic] Fig. 2.1 F= Q·ctgαmin= 8914·ctg30°= 15439.5 N Calculus of the load Fc, N Fc= β·F= 1.3·15439.5= 20071.3 N β= 1.25 ... 1.3 [3] The thread's inner diameter [pic] [pic] [pic]=100 ... 120 Mpa [3] Choosing the thread It is chosen Tr 20X4 with the dimension in table 24.2 Table 2.1 |Nominal diameter |Pitch |Medium diameter |External diameter |Inner diameter | |d, mm |P, mm |d2=D2,, mm |D4, mm | | | | | | | | | | | | | |D3, mm |D1,mm | |20 |4 |18 |20.5 |15.5 |16 | CHECKING THE SELF-BRAKING CONDITION The thread's declination...
Words: 1854 - Pages: 8
...MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Find the x coordinates of all relative extreme points of[pic]. |A)[pic] |B) [pic] |C) [pic] |D) [pic] |E) [pic] | [pic] First find the derivative of the function[pic], f ’(x): |[pic] |= |[pic] |apply power rule of differentiation | | |= |[pic] |simplify | | |= |[pic] |finish simplifying by first factoring | | | | |out GCF | | |= |[pic] |next factor the trinomial factor, | | | | |leaving the final simplified form of | | | | |the derivative | Set[pic]and solve for x to find critical point(s): When the derivative is set to zero, [pic]; thus, this implies each factor could be equal to zero...
Words: 1643 - Pages: 7
...Extension conducting water sampling for effluent matter at local bays. It was also at this point that I started to give thought to what careers I may want to pursue, specifically in a scientific field. In my past three years of high school, I've taken great initiative to enrich my scientific experience and identify which fields directly interest me. I became a member and now president of my school's selective science research program, attended lectures at Stony Brook University and started ready Scientific News. Reading about Physics made me inquisitive about the unknown. In math I started a trail-blazing path by self-teaching Math B (does it have another name) during the summer between 9th and 10th grade which allowed me to go onto Pre-Calculus sophomore year. To prepare myself for my Intel Research project in Astrophysics I decided it would be beneficial to learn Applied Linear Algebra at Stony Brook University. To-date I have completed all possible math...
Words: 654 - Pages: 3
...CHEM 1035 | GENERAL CHEMISTRY | 4 | A | ENGE 1024 | ENGINEERING EXPLORATION | 2 | A | ENGL 1105 | FRESHMAN ENGLISH | 3 | A | MATH 1205 | CALCULUS | 3 | A- | ECON 2005 | MICRO PRINCIPLES OF ECONOMICS | 3 | A- | ACIS 1504 | INTRODUCTION TO BUSINESS INFORMATION SYSTEMS | 3 | B+ | ENGE 1114 | EXPLORATION OF ENGINEERING DESIGN | 2 | A | MATH 1206 | CALCULUS | 3 | B+ | PHYS 2305 | FOUNDATIONS OF PHYSICS | 4 | A | AOE 2074 | COMPUTATIONAL METHODS | 3 | B+ | ECON 2006 | MACRO PRINCIPLES OF ECONOMICS | 3 | A- | ECON 3104 | MICRO ECONOMICS THEORY | 3 | B+ | AOE 2104 | INTRODUCTION TO AEROSPACE ENGINEERING | 3 | A- | ESM 2104 | STATICS | 3 | B | MATH 2224 | MULTIVARIABLE CALCULUS | 3 | A- | AOE 3094 | MATERIALS FOR AEROSPACE AND OCEAN ENGINEERING | 3 | B+ | ACIS 2115 | PRINCIPLES OF ACCOUNTING | 3 | A | BIT 2405 | QUANTITATIVE METHODS | 3 | A | AOE 3104 | AIRCRAFT PERFORMANCE | 3 | B+ | ESM 2204 | MECHANICS OF DEFORMABLE BODIES | 3 | B+ | ESM 2304 | DYNAMICS | 3 | A- | ECON 3204 | MACRO ECONOMICS THEORY | 3 | B+ | MGT 3304 | MANAGEMENT THEORY AND LEADERSHIP PRACTICE | 3 | A | AOE 3054 | AEROSPACE EXPERIMENTAL METHODS | 3 | B+ | AOE 3114 | COMPRESSIBLE AERODYNAMICS | 3 | B | AOE 3124 | AEROSPACE STRUCTURES | 3 | A- | AOE 3134 | STABILITY AND CONTROL | 3 | A | MKT 3104 | MARKETING MANAGEMENT | 3 | B+ | FIN 3104 | INTRODUCTION TO FINANCE | 3 | A | AOE 3044 | BOUNDARY LAYER THEORY | 3 | B | AOE 4154 | AEROSPACE ENGINEERING LAB | 1 |...
Words: 383 - Pages: 2
...Master Dynamique terrestre et risques naturels Math´matiques pour g´ologues e e Op´rateurs diff´rentiels e e On ´tudie en g´osciences des fonctions scalaires des coordonn´es d’espace, comme la temp´rature, e e e e ou bien des vecteurs dont les trois composantes sont des fonctions des coordonn´es, comme la e pesanteur ou le champ magn´tique. Lorsque ces fonctions ont des d´riv´es partielles, on peut e e e d´finir d’autres scalaires ou vecteurs qui restent les mˆmes pour tout r´f´rentiel, ce qu’on appelle e e ee des invariants de la fonction ou du champ de vecteurs. Ils en fournissent des propri´t´s intrins`ques ee e et locales. Pour une fonction, les invariants qui nous seront utiles sont le gradient (un vecteur) et le laplacien (un scalaire). Pour un champ de vecteurs ce sont le rotationnel (un vecteur), la divergence (un scalaire) et le laplacien vectoriel (un vecteur). 1 Produit scalaire et vectoriel Soit deux vecteurs a et b ayant pour composantes dans un r´f´rentiel cart´sien ax , ay , az et bx , by , ee e bz respectivement. On d´finit : e – le produit scalaire : a.b = ax b + ay by + az bz x ay b z − az b y – le produit vectoriel : a ∧ b = az bx − ax bz ax b y − ay b x 2 Notion de circulation d’un champ de vecteurs On appelle travail de A ` B du vecteur a le long d’une courbe (C), dont un segment infinit´simal a e est le vecteur dl, la somme des produits scalaires infinit´simaux e B a.dl. A Lorsque (C) est une courbe ferm´e (A = B), la position...
Words: 1394 - Pages: 6
...Mathematisch-statistische Ansätze zur Aktienkursprognose Seite 1 1 1.1 Einführung Ziel der Arbeit Ziel dieser Arbeit ist die Einordnung, Darstellung, Erläuterung und Bewertung mathematisch-statistischer Verfahren zur Aktienkursprognose. In diesem Zusammenhang werden hierzu neben dem Fokus auf die Prognose von Aktienkursen bzw. -renditen auch die methodologischen Rahmenbedingungen der zugehörigen Finanzmarkttheorie sowie die grundsätzlichen Probleme bei der Anwendung von Prognoseverfahren auf Aktienkurszeitreihen angesprochen. 1.2 Einordnung der Thematik in den aktuellen Forschungsstand Verfahren zur Prognose von Aktienkursen werden schon seit Bestehen von Börsen und anderen Handelsplätzen diskutiert. Somit hat das Thema dieser Arbeit seine ideellen Wurzeln in der von Charles H. Dow begründeten Lable Dow Theorie, die die Technische Aktienanalyse um 1900 begründete. Durch die ab 1965 von Eugene F. Fama proklamierten Thesen informationseffizienter Kapitalmärkte, nach der technische Aktienanalysen wirkungslos sind, erlebte die Kursprognose einen ersten Rückschlag. Die Thematik dieser Arbeit ist der Technischen Aktienanalyse zuzuordnen – nicht zuletzt wurde aber genauso Kritik an den Thesen informationseffizienter Kapitalmärkte geübt, sodass sich diese Antithese in neuerer Zeit verweichlicht hat. Die empirische Kapitalmarktforschung bemüht in letzter Zeit Ansätze des Forschungsgebietes der Behavioral Finance, die versuchen, diese Thesen und real beobachtbare...
Words: 18834 - Pages: 76
...Limit The limx→afx=L means that as x becomes closer and closer to a specific value a, the function f(x) will approach L. L is the number that the value approaches. The function never actually reaches this “a” value, so instead we find what it approaches. Let’s take the function fx=x2-1x-1, at x=1, because it equals 00 so we can take the limit to find what y value, x is approaching. I can either take the limit from the right, or the left. For this example I took the limit as it approached from the right. We get the equation: limx→1+x2-1x-1 = L. Now we can create a table of values where it is approaching x from the right. x | 2 | 1.5 | 1.1 | 1.01 | 1.001 | 1.0001 | f(x) | 3 | 2.5 | 2.1 | 2.01 | 2.001 | 2.0001 | As you can see from the table, as x approaches 1, f(x) approaches 2 but is never actually 2. Since we can’t actually write that f (1)=2 you express it as, limx→1+x2-1x-1=2. The limit does not exist if limx→a-fx≠limx→a+f (x) because in order for the limit to exist, the function must be approaching the “L” value from both the left and the right towards the same number. In the same example above, yes it is possible for this statement f(a)≠L to be true because in in the above example the limit value above is 2 but the function does not exist. Asymptote An asymptote is a value that f(x) approaches when x is approaching an “a” value. In the equation limx→a+fx=∞, this is saying that as x becomes sufficiently close to an “a” value, the function f(x) will approach infinity...
Words: 893 - Pages: 4
...Mathematics learning performance and Mathematics learning difficulties in China Ningning Zhao Promotor: Prof. Dr. Martin Valcke Co-promoter: Prof. Dr. Annemie Desoete Proefschrift ingediend tot het behalen van de academische graad van Doctor in de Pedagogische Wetenschappen 2011 This Ph.D research project was funded by Ghent University BOF Research Grant (BOF07/DOS/056) Acknowledgements There is still a long and indistinct way and I will keep on going to explore the unknown. 路漫漫其修远兮，吾将上下而求索。 - Qu Yuan (340-278 BC) This dissertation would not have been possible unless so many persons contributed to it. The first person I should give my gratitude is Prof. dr. Cong Lixin in Beijing Normal University. It is she who recommanded me to my promoter - Prof. dr. Martin Valcke. Based on the cooperation contact between the two universities, I have the opportunity to start my journey in Ghent University. The fantasty journey started from Year 2007 gudided by the Prof. dr. Martin Valcke. I am heartily thankful to my promoter Prof. dr. Martin Valcke and my co-promotor Prof. dr. Annemie Desoete, whose encouragement, supervision and support from the preliminary to the concluding level enabled me to carry on the research project. My deepest gratitude is to Prof. dr. Martin Valcke. I am not a smart student who always give him so much revision work. It is extremely fortunate for me to have a promoter who is characterized by energy...
Words: 5832 - Pages: 24
...XEQ 201: Calculus II Contents Course description References iv iv Chapter 1. Applications of Differentiation 1.1. Mean value theorems of differential calculus 1.2. Using differentials and derivatives 1.3. Extreme Values iii 1 1 5 7 Course description Application of differentiation. Taylor theorem. Mean Value theorem of differential calculus. Methods of integration. Applications of integration. References 1. Calculus: A complete course by Robert A. Adams and Christopher Essex. 2. Fundamental methods of mathematical economics by Alpha C. Chiang. 3. Schaum’s outline series: Introduction to mathematical economics by Edward T. Dowling iv CHAPTER 1 Applications of Differentiation 1.1. Mean value theorems of differential calculus Theorem 1.1.1 (Mean Value Theorem). Suppose that the function f is continuous on the closed finite interval [a, b] and that it is differentiable on the interval (a, b). Then ∃ a point c ∈ (a, b) such that f (b) − f (a) = f (c) . b−a It means that the slope of the chord joining the points (a, f (a)) and (b, f (b)) is equal to the slope of the tangent line to te curve y = f (x) at the point (c, f (c)) so that the two lines are parallel. Fig 2.28 Example 1.1.1. √ Verify the conclusion of the mean value theorem for f (x) = x on the interval [a, b], where a ≤ x ≤ b. Solution. We are to show that ∃ c ∈ (a, b) such that f (b) − f (a) = f (c) b−a 1 XEQ 201 so long as f is continuous on [a, b] and is...
Words: 2582 - Pages: 11
...Ariana Corney Aesa McComb Mina Haw Son Writing assignment #2 When g (0) it is zero because the slope is at zero When it is g (10) it is positive because of the area being colored on the graph. The area under the x-axis is negative, so when you add all of the areas it becomes positive. That brings us to the derivative of g (x), to help us find all of the x-values, which is equal to f(x). Same with the derivative of g (a), it equals f (a), but f (a) is equal to 0. The x values are 2,4,6,8, and 10 when f (0) is equal to f (2), f (4), f (6), f (8), f (10). We determined the increasing/decreasing intervals would be increasing at points where x equals (0,2), (4,6), (8,10) while that g would be decreasing on points where x equals (2,4) and (6,8). As far as finding the extreme values we found that the maximum occurs at when x equals 2, 6, and 10. The minimum occurs at when x equals 0, 4, and 8, while the absolute extreme values are where x equals 2. We determined that g was concaving up when x equals (0,1), (3,5), and (7,9) because the slope of those are increasing throughout the graph. We also determined that g is concaving down when x equals (1,3), (5,7), (9,10) because the slope of those are decreasing down throughout the graph. When the derivative of h is 9, h(x) =g (√x) is equal to f(3)=-2. H(x)= g(√x) so when we need to find the derivative of h(x), we also need to find the derivative of g(√x).The derivative of h(9) then equals the derivative of g(√9) which equals 3. When...
Words: 352 - Pages: 2
...Game Theory Math Review 1 1.1 Function Definition If we write down the relation of x and y as follows, y = f (x) this means y is related to x under the rule f . Also we say that the value of y depends on the value of x. This relation y = f (x) is called as a function if 1) the rule f assigns a single x value to single y value or 2) assigns multiple x values to single y value. 2 2.1 Shape of function When f (x) = ax + b Suppose that the function is given as follows. y = f (x) = ax + b ´ 1) Slope: f (x) = a and Y −intercept: b Y − axis is b.. b −a 2) Intersection with Y − axis: This is the case when x = 0. So from f (0) = b, the Intersection with 3) Intersection with X − axis: This is the case when y = 0. So the intersection with X − axis is from ax + b = 0. Example Suppose y = f (x) = ax + b. Draw this function in following each case. 1) When a > 0, b > 0 2) When a > 0, b < 0 3) When a < 0, b > 0 4) When a < 0, b < 0 Now suppose you want to find the linear function that passes through following two points,. x = (a, b) and y = (c, d). Then the linear function is defined as follows. ¶ µ b−d (x − a) + b f (x) = a−c µ ¶ b−d = (x − c) + d a−c ´ ³ ´ ³ b−d b−d So from f (x) = a−c (x − a) + b or f (x) = a−c (x − c) + d, ¶ (bc − ad) b−d x+ f (x) = a−c (c − a) | {z } | {z } µ 1 Here, the first term is the slope and the second term is Y-intercept. Example Find the linear function that passes through following two points. A = (2, 4) and B = (−4, −2) 2...
Words: 1496 - Pages: 6
