Graphs of Exponential Functions

Graphs of Exponential Functions tài liệu, giáo án, bài giảng , luận văn, luận án, đồ án, bài tập lớn về tất cả các lĩnh...

Linear recurrences and asymptotic behavior of exponential sums of symmetric boolean functions Francis N. Castro Department of Mathematics University of Puerto Rico, San Juan, PR 00931 francis.castro@upr.edu Luis A. Medina Department of Mathematics University of Puerto Rico, San Juan, PR 00931 luis.medina17@upr.edu Submitted: Jan 28, 2011; Accepted: May 13, 2011; Published: May 25, 2011 Mathematics Subject Classification: 11T23, 05E05 Dedicated to Doron Zeilberger on the occasion of his 60th birthday Abstract In this paper we give an improvement of the degree of the homogeneous linear recurrence with integer coefficients that exponential sums of symmetric Boolean functions satisfy. This improvement is tight. We also compute the asymptotic behavior of symmetric Boolean functions and provide a formula that allows us to determine if a symmetric boolean function is asymptotically not balanced. In par- ticular, when the degree of the symmetric function is a power of two, then the exponential sum is much smaller than 2 n . Keywords: Exponential sums, recurrences, Cusick et al. Conjecture for elementary balanced symmetric boolean functions 1 Introduction Boolean functions are one of the most studied objects in mathematics. They are impor- tant in many applications, for example, in the design of stream ciphers, block and hash functions. These functions also play a vital role in cryptography as they are used as filter and combination generator of stream ciphers based on linear feed-back shift registers. The the electronic journal of combinatorics 18(2) (2011), #P8 1 case of boolean functions of degree 2 has been intensively studied because of its relation to bent functions (see [11], [1]). One can find many papers and books discussing the properties of boolean functions (see [5], [9], [2] and [6]). The subject can be studied from the point of view of complexity theory or from the algebraic point of view as we do in this paper, where we compute the asymptotic behavior of exponential sums of symmetric boolean functions. The correlation between two Boolean functions of n inputs is defined as the number of times the functions agree minus the number of times they disagree all divided by 2 n , i.e., C(F 1 , F 2 ) = 1 2 n  x 1 , ,x n ∈{0,1} (−1) F 1 (x 1 , ,x n )+F 2 (x 1 , ,x n ) . (1.1) In this paper we are interested in the case when F 1 and F 2 are symmetric boolean func- tions. Without loss of generality, we write C(F ) instead of C(F 1 , F 2 ), where F is a symmetric boolean function. In [4], A. Canteaut and M. Videau studied in detail sym- metric boolean functions. They established a link between the periodicity of the simplified value vector of a symmetric Boolean function and its degree. They also determined all balanced symmetric functions of degree less than or equal to 7. In [13], J. von zur Gathen and J. Rouche found all the balanced symmetric boolean functions up to 128 variables. In[3], J. Cai et. al. computed a closed formula for the correlation between any two symmetric Boolean functions. This formula implies that C(F ) satisfies a homogeneous linear recurrence with integer coefficients and provides an upper bound for the degree of the minimal recurrence of this type that C(F) satisfies. In this paper we give an improvement to the degree of the minimal homogeneous linear recurrence with integer coefficients satisfying by C(F ). In particular, our lower and upper bounds are tight in many cases. Also, in the case of an elementary symmetric function we provide the minimal homogeneous linear recurrence. We also compute the asymptotic value of C(F ). In particular, we give infinite families of b oolean functions that are asymptotically not balanced, i.e., lim n→∞ C(F ) = 0. In [7], T. Cusick et al. conjectured that there are no nonlinear balanced elementary symmetric polynomials except for the elementary symmetric boolean function of degree k = 2 r in 2 r · l − 1 variables, where r and l are any positive integers. In this paper, we prove that Graphs of Exponential Functions Graphs of Exponential Functions By: OpenStaxCollege As we discussed in the previous section, exponential functions are used for many realworld applications such as finance, forensics, computer science, and most of the life sciences Working with an equation that describes a real-world situation gives us a method for making predictions Most of the time, however, the equation itself is not enough We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool It gives us another layer of insight for predicting future events Graphing Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth Recall the table of values for a function of the form f(x) = bx whose base is greater than one We’ll use the function f(x) = 2x Observe how the output values in [link] change as the input increases by x −3 −2 −1 f(x) = 2x 2 Each output value is the product of the previous output and the base, We call the base the constant ratio In fact, for any exponential function with the form f(x) = abx, b is the constant ratio of the function This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a Notice from the table that • the output values are positive for all values of x; • as x increases, the output values increase without bound; and • as x decreases, the output values grow smaller, approaching zero 1/33 Graphs of Exponential Functions [link] shows the exponential growth function f(x) = 2x Notice that the graph gets close to the x-axis, but never touches it The domain of f(x) = 2x is all real numbers, the range is (0, ∞), and the horizontal asymptote is y = To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form f(x) = bx whose base is between zero and one We’ll use the function g(x) = increases by x g(x) = x ( 12 ) Observe how the output values in [link] change as the input −3 −2 −1 ( 12 ) x 1 2/33 Graphs of Exponential Functions Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio Notice from the table that • the output values are positive for all values of x; • as x increases, the output values grow smaller, approaching zero; and • as x decreases, the output values grow without bound [link] shows the exponential decay function, g(x) = The domain of g(x) = asymptote is y = ( 12 ) x x ( 12 ) is all real numbers, the range is (0, ∞), and the horizontal A General Note 3/33 Graphs of Exponential Functions Characteristics of the Graph of the Parent Function f(x) = bx An exponential function with the form f(x) = bx, b > 0, b ≠ 1, has these characteristics: • • • • • • • • one-to-one function horizontal asymptote: y = domain: ( – ∞, ∞) range: (0, ∞) x-intercept: none y-intercept: (0, 1) increasing if b > decreasing if b < [link] compares the graphs of exponential growth and decay functions How To Given an exponential function of the form f(x) = bx, graph the function Create a table of points Plot at least point from the table, including the y-intercept (0, 1) Draw a smooth curve through the points 4/33 Graphs of Exponential Functions State the domain, ( − ∞, ∞), the range, (0, ∞), and the horizontal asymptote, y = Sketching the Graph of an Exponential Function of the Form f(x) = bx Sketch a graph of f(x) = 0.25x State the domain, range, and asymptote Before graphing, identify the behavior and create a table of points for the graph • Since b = 0.25 is between zero and one, we know the function is decreasing The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = • Create a table of points as in [link] x −3 −2 −1 f(x) = 0.25x 64 16 0.25 0.0625 0.015625 • Plot the y-intercept, (0, 1), along with two other points We can use ( − 1, 4) and (1, 0.25) Draw a smooth curve connecting the points as in [link] The domain is ( − ∞, ∞); the range is (0, ∞); the horizontal asymptote is y = Try It 5/33 Graphs of Exponential Functions Sketch the graph of f(x) = 4x State the domain, range, and asymptote The domain is ( − ∞, ∞); the range is (0, ∞); the horizontal asymptote is y = Graphing Transformations of Exponential Functions Transformations of exponential graphs behave similarly to those of other functions Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function f(x) = bx without loss of shape For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied Graphing a Vertical Shift The first transformation occurs when we add a constant d to the parent function ...Volume 10 (2009), Issue 3, Article 72, 5 pp. SOLUTION OF ONE CONJECTURE ON INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS LADISLAV MATEJÍ ˇ CKA INSTITUTE OF INFORMATION ENGINEERING, AUTOMATION AND MATHEMATICS FACULTY OF CHEMICAL FOOD TECHNOLOGY SLOVAK UNIVERSITY OF TECHNOLOGY IN BRATISLAVA SLOVAKIA matejicka@tnuni.sk Received 20 July, 2009; accepted 24 August, 2009 Communicated by S.S. Dragomir ABSTRACT. In this paper, we prove one conjecture presented in the paper [V. Cîrtoaje, On some inequalities with power-exponential functions, J. Inequal. Pure Appl. Math. 10 (2009) no. 1, Art. 21. http://jipam.vu.edu.au/article.php?sid=1077]. Key words and phrases: Inequality, Power-exponential functions. 2000 Mathematics Subject Classification. 26D10. 1. INTRODUCTION In the paper [1], V. Cîrtoaje posted 5 conjectures on inequalities with power-exponential functions. In this paper, we prove Conjecture 4.6. Conjecture 4.6. Let r be a positive real number. The inequality (1.1) a rb + b ra ≤ 2 holds for all nonnegative real numbers a and b with a + b = 2, if and only if r ≤ 3. 2. PROOF OF CONJECTURE 4.6 First, we prove the necessary condition. Put a = 2 − 1 x , b = 1 x , r = 3x for x > 1. Then we have (2.1) a rb + b ra > 2. In fact,  2 − 1 x  3 +  1 x  3x ( 2− 1 x ) = 8 − 12 x + 6 x 2 − 1 x 3 +  1 x  6x−3 The author is deeply grateful to Professor Vasile Cîrtoaje for his valuable remarks, suggestions and for his improving some inequalities in the paper. 193-09 2 LADISLAV MATEJÍ ˇ CKA and if we show that  1 x  6x−3 > −6 + 12 x − 6 x 2 + 1 x 3 then the inequality (2.1) will be fulfilled for all x > 1. Put t = 1 x , then 0 < t < 1. The inequality (2.1) becomes t 6 t > t 3 (t 3 − 6t 2 + 12t − 6) = t 3 β(t), where β(t) = t 3 − 6t 2 + 12t − 6. From β  (t) = 3(t − 2) 2 , β(0) = −6, and from that there is only one real t 0 = 0.7401 such that β(t 0 ) = 0 and we have that β(t) ≤ 0 for 0 ≤ t ≤ t 0 . Thus, it suffices to show that t 6 t > t 3 β(t) for t 0 < t < 1. Rewriting the previous inequality we get α(t) =  6 t − 3  ln t − ln(t 3 − 6t 2 + 12t − 6) > 0. From α(1) = 0, it suffices to show that α  (t) < 0 for t 0 < t < 1, where α  (t) = − 6 t 2 ln t +  6 t − 3  1 t − 3t 2 − 12t + 12 t 3 − 6t 2 + 12t − 6 . α  (t) < 0 is equivalent to γ(t) = 2 ln t − 2 + t + t 2 (t − 2) 2 t 3 − 6t 2 + 12t − 6 > 0. From γ(1) = 0, it suffices to show that γ  (t) < 0 for t 0 < t < 1, where γ  (t) = (4t 3 − 12t 2 + 8t)(t 3 − 6t 2 + 12t − 6) − (t 4 − 4t 3 + 4t 2 )(3t 2 − 12t + 12) (t 3 − 6t 2 + 12t − 6) 2 + 2 t + 1 = t 6 − 12t 5 + 56t 4 − 120t 3 + 120t 2 − 48t (t 3 − 6t 2 + 12t − 6) 2 + 2 t + 1. γ  (t) < 0 is equivalent to p(t) = 2t 7 − 22t 6 + 92t 5 − 156t 4 + 24t 3 + 240t 2 − 252t + 72 < 0. From p(t) = 2(t − 1)(t 6 − 10t 5 + 36t 4 − 42t 3 − 30t 2 + 90t − 36), it suffices to show that (2.2) q(t) = t 6 − 10t 5 + 36t 4 − 42t 3 − 30t 2 + 90t − 36 > 0. Since q(0.74) = 5.893, q(1) = 9 it suffices to show that q  (t) < 0 and (2.2) will be proved. Indeed, for t 0 < t < 1, we have q  (t) = 2(15t 4 − 100t 3 + 216t 2 − 126t − 30) < 2(40t 4 − 100t 3 + 216t 2 − 126t − 30) = 4(t − 1)(20t 3 − 30t 2 + 78t + 15) < 4(t − 1)(−30t 2 + 78t) < 0. This completes the proof of the necessary condition. We prove the sufficient condition. Put a = 1 − x and b = 1 + x, where 0 < x < 1. Since the desired inequality is true for x = 0 and for x = 1, we only need to show that (2.3) (1 − x) r(1+x) + (1 + x) r(1−x) ≤ 2 for 0 < x < 1, 0 < r ≤ 3. Denote ϕ(x ) = (1− x) r(1+x) +(1 + x) r(1−x) . We show that ϕ  (x) < 0 for 0 < x < 1, 0 < r ≤ 3 which gives that (2.3) is valid (ϕ(0) = 2). ϕ  (x) = (1 − x) r(1+x)  r ln(1 − x) − r 1 + x 1 − x  + (1 + x) r(1−x)  r 1 − x 1 + x − r ln(1 + x)  . J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 72, 5 pp. http://jipam.vu.edu.au/ SOLUTION OF ONE CONJECTURE 3 The inequality ϕ  (x) < 0 is equivalent to (2.4)  1 + x 1 − x  r  1 − x 1 + LOCAL POLYNOMIAL CONVEXITY OF GRAPHS OF FUNCTIONS IN SEVERAL VARIABLES KIEU PHUONG CHI Abstract. In that paper, we investigate the locally polynomial convexity of graphs of smooth functions in several variables. We also give a sufficient condition for real analytic function g defined near 0 in C which behaves like z n near the origin so that the algebra generated by z m and g is dense in the space of continuous functions on D for all disks D close enough to the origin in C. 1. Introduction ˆ we denote the polynomial convex We recall that for a given compact K in Cn , by K hull of K i.e., ˆ = {z ∈ Cn : |p(z)| ≤ p K K for every polynomial p in Cn }. ˆ = K. A compact K is called locally We say that K is polynomially convex if K polynomially convex at a ∈ K if there exists the closed ball B(a) centered at a such that B(a) ∩ K is polynomially convex. The interest for studying polynomial convexity stems from the celebrated Oka-Weil approximation theorem (see [1], page 36) which states that holomorphic functions near a compact polynomially convex subset of Cn can be uniformly approximated by polynomials in Cn . A compact K ⊂ C is polynomially convex if is C \ K connected. In higher dimensions, there is no such topological characterization of polynomially convex sets, and it is usual difficult to determine whether a given compact subset is polynomially convex. By a well-known result of Wermer ([19]; see also [1], Theorem. 17.1), every totally real manifold is locally polynomially convex. Recall that a C 1 smooth real manifold M is called totally real at p ∈ M if the real tangent space Tp M contains no complex line. In this paper, we are concerned with local polynomial convexity at the origin of the graph Γf of a C 2 smooth function f near 0 ∈ Cn such that f (0) = 0. By the theorem of Wermer just cited, we ∂f know that if (0) = 0 for all i = 1, 2, ..., n then Γf is locally polynomially convex at ∂z i ∂f the origin of Cn+1 . Thus it remains to consider the case where (0) = 0 for some i. ∂z i Our study is motivated by a similar problem in one complex variable. More precisely, let f be a C 2 smooth function near 0 ∈ C such that f (0) = 0. Under certain condition of f , one can show that Γf is locally polynomial convex at the origin of C2 . The work 2010 Mathematics Subject Classification. 46J10, 46J15, 47H10. Key words and phrases. polynomially convex, plurisubharmonic, totally real . 1 associated with these direction of research is too numerous to list here; instead, the reader is referred to [2, 3, 20] and the references given therein. In the section 3, we will refine the technical from [6] to attack the problem in several variables. For the readers convenience, we repeat a reasoning due to [6]. First, we construct nonnegative smooth functions vanishing exactly on Γf . These functions are, in general, plurisubharmonic only on open sets whose boundaries contain the origin. Secondly,under some technical assumptions, we may add small strictly plurisubharmonic functions to obtain plurisubharmonic functions on certain open sets containing the (local) polynomially convex hull of Γf . Finally, by invoking the nontrivial fact of about equivalence of plurisubharmonic hull and polynomial hulls, we can conclude that Γf is locally polynomially convex at the origin. In this vein, we obtain some known results in one variable. We also give some examples to show that our results are effective. In section 4, we shall present some results about locally uniform approximation of continuous function. Let D be a small closed disk in the complex plane, centered at the origin and g be a C 2 function on D which behaves like z n near the origin. By [z m , g; D] we denote the function algebra consisting of uniform limits on D of all polynomials in z m and g. Our goal finding conditions on g Recognizing circulant graphs of prime order in polynomial time ∗ Mikhail E. Muzychuk Netanya Academic College 42365 Netanya, Israel mikhail@netvision.net.il Gottfried Tinhofer Technical University of Munich 80290 M¨unchen, Germany gottin@mathematik.tu-muenchen.de Submitted: December 19, 1997; Accepted: April 1, 1998 Abstract A circulant graph G of order n is a Cayley graph over the cyclic group Z n . Equivalently, G is circulant iff its vertices can be ordered such that the cor- responding adjacency matrix becomes a circulant matrix. To each circulant graph we may associate a coherent configuration A and, in particular, a Schur ring S isomorphic to A. A can be associated without knowing G to be circu- lant. If n is prime, then by investigating the structure of A either we are able to find an appropriate ordering of the vertices proving that G is circulant or we are able to prove that a certain necessary condition for G being circulant is violated. The algorithm we propose in this paper is a recognition algorithm for cyclic association schemes. It runs in time polynomial in n. MR Subject Number: 05C25, 05C85, 05E30 Keywords: Circulant graph, cyclic association scheme, recognition algorithm ∗ The work reported in this paper has been partially supported by the German Israel Foundation for Scientific Research and Development under contract # I-0333-263.06/93 the electronic journal of combinatorics 3 (1996), #Rxx 2 1 Introduction The graphs considered in this paper are of the form (X, γ), where X is a finite set and γ is a binary relation on X which is not necessarily symmetric. Let G be a group and G =(X, γ) a graph with vertex set X = G and with adjacency relation γ defined with the aid of some subset C ⊂Gby γ = {(g, h):g,h ∈G∧gh −1 ∈ C}. Then G is called Cayley graph over the group G. Let Z n , n ∈ N, stand for a cyclic group of order n written additively. A circulant graph G over Z n is a Cayley graph over this group. In this particular case, the adjacency relation γ has the form γ = n−1  i=0 {i}×{i+γ(0)} where γ(0) is the set of successors of the vertex 0. Evidently, the set of successors γ(i) of an arbitrary vertex i satisfies γ(i)=i+γ(0). The set γ(0) is called the connection set of the circulant graph G. G is a simple undirected graph if 0 ∈ γ(0) and j ∈ γ(0) implies −j ∈ γ(0). There are different equivalent characterizations of circulant graphs. One of them is this: A graph G is a circulant graph iff its vertex set can be numbered in such a way that the resulting adjacency matrix A(G) is a circulant matrix. We call such a numbering a Cayley numbering. Still another characterization is: G is a circulant graph iff a cyclic permutation of its vertices exists which is an automorphism of G. Cayley graphs, and in particular, circulant graphs have been studied intensively in the literature. These graphs are easily seen to be vertex transitive. In the case of a prime vertex number n circulant graphs are known to be the only vertex transitive graphs. Because of their high symmetry, Cayley graphs are ideal models for commu- nication networks. Routing and weight balancing is easily done on such graphs. Assume that a graph G on the set V (G)={0, ,n−1} is given by its diagram or by its adjacency matrix, or by some other data structure commonly used in dealing with graphs. How can we decide whether G is a Cayley graph or not? In such a generality, this decision problem seems to be far from beeing tractable efficiently. A recognition algorithm for Cayley graphs would have to involve implicitly checking all finite groups of order n. In the special case of circulant graphs, or in any other case where the group G is given, we could recognize Cayley graphs by checking all different numberings of the vertex set and comparing [...]... Departments at the offices / branches of the Reserve Bank, while management of public debt including floatation of new loans is done at Public Debt Office at offices / branches of the Reserve Bank and by the Internal Debt Management Department at the Central Office For the final compilation of the Government accounts, both of the centre and states, the Nagpur office of the Reserve Bank has a Central... 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Board of Directors The Central Board of Directors is at the top of the Reserve Bank s organisational structure Appointed by the Government under the provisions of the Reserve Graphs of Logarithmic Functions Graphs of Logarithmic Functions By: OpenStaxCollege In Graphs of Exponential Functions, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events How logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation In other words, logarithms give the cause for an effect To illustrate, suppose we invest $2500 in an account that offers an annual interest rate of 5%, compounded continuously We already know that the balance in our account for any year t can be found with the equation A = 2500e0.05t But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation By graphing the model, we can see the output (year) for any input (account balance) For instance, what if we wanted to know how many years it would take for our initial investment to double? [link] shows this point on the logarithmic graph 1/45 Graphs of Logarithmic Functions In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions Finding the Domain of a Logarithmic Function Before working with graphs, we will take a look at the domain (the set of input values) for which ... Transformations of Exponential Functions Transformations of exponential graphs behave similarly to those of other functions Just as with other parent functions, we can apply the four types of transformations—shifts,... transformation of y = 2x Write an equation describing the transformation 29/33 Graphs of Exponential Functions y = − 2x + 30/33 Graphs of Exponential Functions For the following exercises, find an exponential. .. f(x) = (4) x 26/33 Graphs of Exponential Functions x f(x) = 3(0.75) − x f(x) = − 4(2) + 27/33 Graphs of Exponential Functions For the following exercises, graph the transformation of f(x) = 2x Give