Exponential Functions

Text/String Functions codepoint-equal(xs:string?, xs:string?) as xs:boolean? codepoints-to-string(xs:integer*) as xs:string compare(xs:string?, xs:string?) as xs:integer? compare(xs:string?, xs:string?, xs:string) as xs:integer? concat(xs:anyAtomicType?, xs:anyAtomicType?, ) as xs:string contains(xs:string?, xs:string?) as xs:boolean contains(xs:string?, xs:string?, xs:string) as xs:boolean current-date() as xs:date current-dateTime() as xs:dateTime current-time() as xs:time default-collation() as xs:string encode-for-uri(xs:string?) as xs:string ends-with(xs:string?, xs:string?) as xs:boolean ends-with(xs:string?, xs:string?, xs:string) as xs:boolean escape-html-uri(xs:string?) as xs:string lower-case(xs:string?) as xs:string normalize-space() as xs:string normalize-space(xs:string?) as xs:string normalize-unicode(xs:string?) as xs:string normalize-unicode(xs:string?, xs:string) as xs:string starts-with(xs:string?, xs:string?) as xs:boolean starts-with(xs:string?, xs:string?, xs:string) as xs:boolean string() as xs:string string(item()?) as xs:string string-join(xs:string*, xs:string) as xs:string string-length() as xs:integer string-length(xs:string?) as xs:integer string-to-codepoints(xs:string?) as xs:integer* substring(xs:string?, xs:double) as xs:string substring(xs:string?, xs:double, xs:double) as xs:string substring-after(xs:string?, xs:string?) as xs:string substring-after(xs:string?, xs:string?, xs:string) as xs:string substring-before(xs:string?, xs:string?) as xs:string substring-before(xs:string?, xs:string?, xs:string) as xs:string translate(xs:string?, xs:string, xs:string) as xs:string upper-case(xs:string?) as xs:string XSL-List: http://www.mulberrytech.com/xsl/xsl-list REGEX Functions matches(xs:string?, xs:string) as xs:boolean matches(xs:string?, xs:string, xs:string) as xs:boolean replace(xs:string?, xs:string, xs:string) as xs:string replace(xs:string?, xs:string, xs:string, xs:string) as xs:string tokenize(xs:string?, xs:string) as xs:string* tokenize(xs:string?, xs:string, xs:string) as xs:string* Arithmetic Operators + (numeric) as ~numeric (numeric) + (numeric) as ~numeric - (numeric) as ~numeric (numeric) - (numeric) as ~numeric (numeric) * (numeric) as ~numeric (numeric) div (numeric) as ~numeric (numeric) idiv (numeric) as xs:integer (numeric) mod (numeric) as ~numeric Arithmetic Functions abs(numeric?) as ~numeric? avg(xs:anyAtomicType*) as ~xs:anyAtomicType? ceiling(numeric?) as ~numeric? floor(numeric?) as ~numeric? number() as xs:double number(xs:anyAtomicType?) as xs:double round(numeric?) as ~numeric? round-half-to-even(numeric?) as ~numeric? round-half-to-even(numeric?, xs:integer) as ~numeric? sum(xs:anyAtomicType*) as ~xs:anyAtomicType sum(xs:anyAtomicType*, xs:anyAtomicType?) as ~xs:anyAtomicType? The eq, ne, lt, gt, le and ge comparisons are supported for the numeric types. Sequence Operators (item()*) , (item()*) as ~item()* (node()*) union (node()*) as ~node()* (node()*) intersect (node()*) as ~node()* (node()*) except (node()*) as ~node()* (xs:integer) to (xs:integer) as xs:integer* Node Comparisons (node()) is (node()) as xs:boolean (node()) << (node()) as xs:boolean (node()) >> (node()) as xs:boolean Sequence and Node Functions collection() as node()* collection(xs:string?) as node()* count(item()*) as xs:integer data(item()*) as ~xs:anyAtomicType* deep-equal(item()*, item()*) as xs:boolean deep-equal(item()*, item()*, string) as xs:boolean distinct-values(xs:anyAtomicType*) as ~xs:anyAtomicType* distinct-values(xs:anyAtomicType*, xs:string) as ~xs:anyAtomicType* doc(xs:string?) as document-node()? empty(item()*) as xs:boolean exactly-one(item()*) as ~item() exists(item()*) as xs:boolean index-of(xs:anyAtomicType*, xs:anyAtomicType) as xs:integer* index-of(xs:anyAtomicType*, xs:anyAtomicType, xs:string) as xs:integer* insert-before(item()*, xs:integer, item()*) as ~item()* last() as xs:integer nilled(node()?) as xs:boolean? Exponential Functions Exponential Functions By: OpenStaxCollege India is the second most populous country in the world with a population of about 1.25 billion people in 2013 The population is growing at a rate of about 1.2% each year http://www.worldometers.info/world-population/ Accessed February 24, 2014 If this rate continues, the population of India will exceed China’s population by the year 2031.When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly To a mathematician, however, the term exponential growth has a very specific meaning In this section, we will take a look at exponential functions, which model this kind of rapid growth Identifying Exponential Functions When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input For example, in the equation f(x) = 3x + 4, the slope tells us the output increases by each time the input increases by The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people Defining an Exponential Function A Harris Interactivity study found that the number of vegans in the United States doubled from 2009 to 2011 In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs If this rate continues, vegans will make up 10% of the U.S population in 2015, 40% in 2019, and 80% in 2050 Bohanec, Hope "U.S vegan population doubles in only two years." Occupy for Animals http://www.occupyforanimals.org/us-vegan-population-doubles-in-only-twoyears.html 1/33 Exponential Functions What exactly does it mean to grow exponentially? What does the word double have in common with percent increase? People toss these words around errantly Are these words used correctly? The words certainly appear frequently in the media • Percent change refers to a change based on a percent of the original amount • Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time • Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth We will construct two functions The first function is exponential We will start with an input of 0, and increase each input by We will double the corresponding consecutive outputs The second function is linear We will start with an input of 0, and increase each input by We will add to the corresponding consecutive outputs See [link] x f(x) = 2x g(x) = 2x 1 2 4 16 32 10 64 12 From [link] we can infer that for these two functions, exponential growth dwarfs linear growth • Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain • Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain Apparently, the difference between “the same percentage” and “the same amount” is quite significant For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input 2/33 Exponential Functions increased by one For linear growth, the constant additive rate of change over equal increments resulted in adding to the output whenever the input was increased by one The general form of the exponential function is f(x) = abx, where a is any nonzero number, b is a positive real number not equal to • If b > 1, the function grows at a rate proportional to its size • If < b < 1, the function decays at a rate proportional to its size Let’s look at the function f(x) = 2x from our example We will create a table ([link]) to determine the corresponding outputs over an interval in the domain from −3 to x −3 −3 f(x) = 2x = −2 2−2 = −1 2−1 = 2 20 = 21 = 22 = 23 = Let us examine the graph of f by plotting the ordered pairs we observe on the table in [link], and then make a few observations 3/33 Exponential Functions Let’s define the behavior of the graph of the exponential function f(x) = 2x and highlight some its key characteristics the domain is ( − ∞, ∞), the range is (0, ∞), as x → ∞, f(x) → ∞, as x → − ∞, f(x) → 0, f(x) is always increasing, the graph of f(x) will never touch the x-axis because base two raised to any exponent never has the result of zero • y = is the horizontal asymptote • the y-intercept is A General Note Exponential Function • • • • • • For any real number x, an exponential function is a function with the form f(x) = abx 4/33 Exponential Functions where • a is the a non-zero real number ...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 Volume 10 (2009), Issue 1, Article 21, 6 pp. ON SOME INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS VASILE CÎRTOAJE DEPARTMENT OF AUTOMATION AND COMPUTERS UNIVERSITY OF PLOIE ¸STI PLOIESTI, ROMANIA vcirtoaje@upg-ploiesti.ro Received 13 October, 2008; accepted 09 January, 2009 Communicated by F. Qi ABSTRACT. In this paper, we prove the open inequality a ea + b eb ≥ a eb + b ea for either a ≥ b ≥ 1 e or 1 e ≥ a ≥ b > 0. In addition, other related results and conjectures are presented. Key words and phrases: Power-exponential function, Convex function, Bernoulli’s inequality, Conjecture. 2000 Mathematics Subject Classification. 26D10. 1. INTRODUCTION In 2006, A. Zeikii posted and proved on the Mathlinks Forum [1] the following inequality (1.1) a a + b b ≥ a b + b a , where a and b are positive real numbers less than or equal to 1. In addition, he conjectured that the following inequality holds under the same conditions: (1.2) a 2a + b 2b ≥ a 2b + b 2a . Starting from this, we have conjectured in [1] that (1.3) a ea + b eb ≥ a eb + b ea for all positive real numbers a and b. 2. MAIN RESULTS In what follows, we will prove some relevant results concerning the power-exponential in- equality (2.1) a ra + b rb ≥ a rb + b ra for a, b and r positive real numbers. We will prove the following theorems. Theorem 2.1. Let r, a and b be positive real numbers. If (2.1) holds for r = r 0 , then it holds for any 0 < r ≤ r 0 . 280-08 2 VASILE CÎRTOAJE Theorem 2.2. If a and b are positive real numbers such that max{a, b} ≥ 1, then (2.1) holds for any positive real number r. Theorem 2.3. If 0 < r ≤ 2, then (2.1) holds for all positive real numbers a and b. Theorem 2.4. If a and b are positive real numbers such that either a ≥ b ≥ 1 r or 1 r ≥ a ≥ b, then (2.1) holds for any positive real number r ≤ e. Theorem 2.5. If r > e, then (2.1) does not hold for all positive real numbers a and b. From the theorems above, it follows that the inequality (2.1) continues to be an open problem only for 2 < r ≤ e and 0 < b < 1 r < a < 1. For the most interesting value of r, that is r = e, only the case 0 < b < 1 e < a < 1 is not yet proved. 3. PROOFS OF THEOREMS Proof of Theorem 2.1. Without loss of generality, assume that a ≥ b. Let x = ra and y = rb, where x ≥ y. The inequality (2.1) becomes (3.1) x x − y x ≥ r x−y (x y − y y ). By hypothesis, x x − y x ≥ r x−y 0 (x y − y y ). Since x − y ≥ 0 and x y − y y ≥ 0, we have r x−y 0 (x y − y y ) ≥ r x−y (x y − y y ), and hence x x − y x ≥ r x−y 0 (x y − y y ) ≥ r x−y (x y − y y ).  Proof of Theorem 2.2. Without loss of generality, assume that a ≥ b and a ≥ 1. From a r(a−b) ≥ b r(a−b) , we get b rb ≥ a rb b ra a ra . Therefore, a ra + b rb − a rb − b ra ≥ a ra + a rb b ra a ra − a rb − b ra = (a ra − a rb )(a ra − b ra ) a ra ≥ 0, because a ra ≥ a rb and a ra ≥ b ra .  Proof of Theorem 2.3. By Theorem 2.1 and Theorem 2.2, it suffices to prove (2.1) for r = 2 and 1 > a > b > 0. Setting c = a 2b , d = b 2b and s = a b (where c > d > 0 and s > 1), the desired inequality becomes c s − d s ≥ c − d. In order to prove this inequality, we show that (3.2) c s − d s > s(cd) s−1 2 (c − d) > c − d. The left side of the inequality in (3.2) is equivalent to f(c) > 0, where f(c) = c s − d s − s(cd) s−1 2 (c − d). We have f  (c) = 1 2 sc s−3 2 g(c), where g(c) = 2c s+1 2 − (s + 1)cd s−1 2 + (s − 1)d s+1 2 . Since g  (c) = (s + 1)  c s−1 2 − d s−1 2  > 0, g(c) is strictly increasing, g(c) > g(d) = 0, and hence f  (c) > 0. Therefore, f (c) is strictly increasing, and then f(c) > f (d) = 0. J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 21, 6 pp. http://jipam.vu.edu.au/ INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS 3 The right side of the inequality in (3.2) is equivalent to a b (ab) a−b > 1. Write this inequality as f(b) > 0, where f(b) = 1 + a − b 1 − a + b ln a − ln b. In order to prove that f(b) > 0, it suffices to show that f  (b) < 0 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 + Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleyChapter 6I/O Streams as an Introduction to Objects and Classes Slide 6- 3Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleyOverview6.1 Streams and Basic File I/O 6.2 Tools for Stream I/O6.3 Character I/O6.4 Inheritance Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley6.1Streams and Basic File I/O Slide 6- 5Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleyI/O StreamsI/O refers to program input and outputInput is delivered to your program via a stream objectInput can be fromThe keyboardA fileOutput is delivered to the output device via a streamobjectOutput can be to The screenA file Slide 6- 6Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleyObjectsObjects are special variables thatHave their own special-purpose functionsSet C++ apart from earlier programming languages Slide 6- 7Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleyStreams and Basic File I/OFiles for I/O are the same type of files used tostore programsA stream is a flow of data.Input stream: Data flows into the programIf input stream flows from keyboard, the program willaccept data from the keyboardIf input stream flows from a file, the program will acceptdata from the fileOutput stream: Data flows out of the programTo the screenTo a file Slide 6- 8Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesleycin And cout StreamscinInput stream connected to the keyboardcout Output stream connected to the screencin and cout defined in the iostream libraryUse include directive: #include <iostream>You can declare your own streams to use with files. Slide 6- 9Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleyWhy Use Files?Files allow you to store data permanently!Data output to a file lasts after the program endsAn input file can be used over and overNo typing of data again and again for testingCreate a data file or read an output file at yourconvenienceFiles allow you to deal with larger data sets Slide 6- 10Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleyFile I/OReading from a fileTaking input from a fileDone from beginning to the end (for now)No backing up to read something again (OK to start over)Just as done from the keyboardWriting to a fileSending output to a fileDone from beginning to end (for now)No backing up to write something again( OK to start over)Just as done to the screen [...]... only to the stream named in the call Slide 6- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Overview 6.1 Streams and Basic File I/O 6.2 Tools for Stream I/O 6.3 Character I/O 6.4 Inheritance Slide 6- 45 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Manipulators  A manipulator is a function called Introduction to Exponential and Logarithmic Functions Introduction to Exponential and Logarithmic Functions By: OpenStaxCollege 1/3 Introduction to Exponential and Logarithmic Functions Electron micrograph of E.Coli bacteria (credit: “Mattosaurus,” Wikimedia Commons) Focus in on a square centimeter of your skin Look closer Closer still If you could look closely enough, you would see hundreds of thousands of microscopic organisms They are bacteria, and they are not only on your skin, but in your mouth, nose, and even your intestines In fact, the bacterial cells in your body at any given moment ... instruction and practice with exponential functions 23/33 Exponential Functions • Exponential Growth Function • Compound Interest Key Equations definition of the exponential function f(x) = bx,... 1095.6 − 2x g(x) = 0.875x andj(x) = 1095.6 − 2x represent exponential functions Evaluating Exponential Functions Recall that the base of an exponential function must be a positive real number other... exponential decay function 13/33 Exponential Functions The graph of f(x) = 2.4492(0.6389)x models exponential decay Try It Given the two points (1, 3) and (2, 4.5), find the equation of the exponential