{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 13 - M.1 The exponential and logarithm functions In...

Info icon This preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: M.1 The exponential and logarithm functions In this section, we study the exponential and logarithm functions and derive their pr0perties. We also define ab for a > 0 and b arbitrary, and we verify the laws of expo- nents. As we did for the trig functions, we shall assume a theorem concerning the existence of the exponential function, postponing the proof until after we have studied power series. Thus we assume the following: Theorem 1. There exists a function E(x), defined fo_r £1 £1 numbers 1:, satisfn'ng th_e following conditions E’(x) = E(x); E(O) = 1. We call E the mnential mm, for reasons to be seen. It is sometimes denoted exp(x). I Theorem 2. (i) The equation E(a+b) = E(a)E(b) holds for all a and b. In particular, E(a)E(—a) = 1 for all a. I (ii) E(x) is continuous, positive, and strictly increasing. (iii) The conditions E’(::) = E(x) and E(O) = .1 determine E(x) uniquely. (iv) If n is an integer and a is a real number, then E(na) = E(a)n. In particular, if e is defined by the equation e = E(l), then E(n) = en. This equation shows why E is called the "exponential function". Rd} (v) The number e satisfies the inequalities 2 5 e 5 4. (vi) E(x) takes on every positive real value exactly once. 2.1M (i) For fixed b, let us set f(x) = E(x+b)E(—x). Then ((x) = E’(x+b)E(—x) — E(x+b)E'(—x) = E(x+b)E(—x) — E(x+b)E(—x) =(l Hence f equals a constant K. Setting x = 0, we see that K = E(b). Thus (*) ‘ E(x+b)E(—x) = mm for all x and b. If we set b = 0 in equation (*), we obtain the equation E(x)E(-x) = 1, which holds for all x. If then we multiply both sides of equation (*) by E(x), we obtain the equation E(x)E(x+b)E(-x) = E(x)E(b), or E(x+b) = E(x)E(b). Setting x = 3. gives our desired equation. (ii) E(x) is continuous because it is differentiable. The equation E(x)E(-—x) = 1 intermediate implies that E(x) 3‘ 0 for all x. The ' A -value theorem then applies to show that, since E(x) is positive for x = 0, it is positive for all x. It follows that E'(x) = E(x) is positive for all x, so that E is strictly increasing. M.3 (iii) Let E(x) be another function satisfying the given conditions. Set . 300 = E(x)E(-x). One checks readily that g'(x) = 0, so that g(x) is constant. Setting x = 0, we see this constant is 1. Hence E(x)E(—x) = 1, or E(x) = E(x). ' (iv) One proves the result for positive integers by induction: The equation E(na)= E(a)n holds for n = 1, trivially. If it holds for n , compute E((n+1)a) = E(na+a) = E(na)E(a) by (i) = E(a)nE(a) by the induction hypothesis, = E(a)n+1. The equation holds when n = 0 by definition (both sides equal 1), and it holds for nega— tive integers because E(na) E(-na) = 1. so that E(—na) = 1/E(na) = 1/E(a)n. (v) Because E is increasing, E(fl : 1 for x 2 0. The comparison theorem implies that 1 5); E(x)dx = I: E’(x)dx = E(1) — E(O) = e— 1. Hence e 2 2. We leave the other inequality as an exercise. (vi) It follows from what we have proved that E(n) = en 2 2n, and E(—n) = 1/E(n) 5 U2“. Given any positive real number r, we may choose a positive integer n such that M.4 n 11 intermediate _ _ 1/2 3 r g 2 . The A—value theorem then implies that E(x) takes on the value I for some x in the interval [—n,n]. '1: Remark. Sinoe en = E(n) for all integers n, it seems reasonable to define ex for arbitrary real x by the equation ex = E(x). Theorems 1 and 2 can then be restated in this new notation, which is standard, as follows: g§(ex)=ex andJexdx=ex+C ea+b = ea _ eb, n ena _ (ea) . The latter two equations are special cases of the laws of exponents, which shall prove shortly in full generality. Remark. The preceding theorem implies that the function E(x) = ex has the following familiar graph. (It is concave upwards because E'(x) = E(x) > 0.) _1. Show that $513” E(x)dxs§Jé. Show the integral equals J6 - 1; conclude that 2.25 5 e 5 4. 2.. Show more generally, by integrating E(x) over the interval [0,1 /n], that 1/n 3 “J6 — 1 5 “film Conclude that 1 n 1 n+1 (1+3) S e S (1+3) M.5 m='-I- These inequalities give a (not very useful) way of computing e. Try n = 2 and 3:3 in these formulas, using your calculator. Definition. The function ex is strictly increasing and takes on each positive value exactly once. We define the (natural) logarithm {firm to be its inverse. That is, if y is any positive number, we define Iogy=x ifandonlyif y=ex. The logarithm. function thus has the graph 7L It is strictly increasing and continuous. It is defined only for y > 0, and it takes on every real value. The fact that these functions are inverses of each other implies that: elogyzy if y>0, log(ex) = x for all x. ThQrem a. The logarithm function has the following properties: (i) g—iabg x) =§ and Hdniog Ix] + C. (ii) log (ab) = log a. + log b if a and b are positive. (iii) log(an) = n log a if n is an integer and a is positive. M (i) Let f(x) = ex and g(y) = log y. Then g is the inverse function to f. We use the formula for the derivative of an inverse function: M.6 , g’(y) may, Now f (x) = f(x) for all x, so g’(y) mgmwog—fig since f(s(y)) = el°g Y = y. If x is positive, the derivative of log|x| = log x is 1/x. If x is negative, the derivative of log |x| = log(—x) is (—1)/(—-x) = l/x. Thus I dx/x = loglxl + C. (ii) Given a and b, let x=loga and y=lOgb. We have the equation . e’H'y = ex - ey. Since both sides of this equation are positive, we can take their logs to conclude that x + y = log(ex-ey), so that I log a. + 10gb = log(a-b), _ as desired. (iv) Given a > 0, let x = log a. Then a = ex, so 11 an = (ex) = ex1x by (iv) of Theorem 2, so that 103 an = nx = n(log a). U Thgrem 4. There is one and only one function ax, defined for all positive a. and all real x, such that the following four conditions hold: (i) The function a1‘ is positive and continuous. M.7 (ii) a1 = a. (iii) ax+y = axay. (iv) (ax)y = axy. This function satisfies the following additional conditions: (v) axbx = (ab)x. (vi) log ax = x log a. 31-29;. Uniqueness. Suppose ax is defined and satisfies conditions (i) — (iv). Conditions (ii) and (iii) imply that a1: a and an+1 = all-a for every positive integer n. The equation a141.0 = a1+0 = a1 implies (since a > 0) that a0 = 1. Finally, the equation ana-n = a0 = 1 implies that a"1 = l/an. Hence integral powers of a iant be defined as we have defined them earlier. Now if n is a positive integer and m is any integer, (iv) implies that (9/31)" = a or al/n = 91’; . Then, using (iv) again, we see that (am/n) = (’15 )m Thus for x rational, ax is completely determined by conditions (ii) — (iv) and positivity. (al/n)m Continuity now implies that ax is determined for all 1:: Suppose f(x) and g(x) are two functions that satisfy (i)—-(iv). Let x0 be arbitrary. Given 5 > 0, choose 6 so that M.8 |f(x)—f(x0)| < e and |g(x)—g(x0)| < e for Ix—xol < 6. Then choose x1 rational with [xl—xol < 6. It follows from what we have already showed that f(x1) = g(x1). Then |f(x0)—g(x0)| < 26. Since 6 is arbitrary, we must have {(3:0) = g(x0). Existence. We motivate the definition as follows: -If n is a positive integer, then 11 n n Ioga= log( J5) = nlog (5, so that ‘n m n m log( J5) =mlog J5=floga, or (“mm = E(%1og a). This equation suggests the following definition. We define, for arbitrary x, ax = E(x log a). Then condition (vi) holds trivially, for log ax = x log a by definition. _ We show that the other conditions of the theorem are satisfied: (i) a1‘ is positive and continuous. (ii) a.1 = E(log a) = a. (iii) ax” = E((x+y) log a) by definition, = E(x log a+ y log a) by distributivity, = E(x log a) - E(y log a) by Theorem 2, = a. - ay. by definition. M.9 (iv) (ax)y = E(y log ax) by definition, = E(Y(x log a)) by (vi), = E((xy) log a) by associativity, = axy. by definition. (v) (ab)x = E(x log (ab)) by definition, = E(x(log a + log b)) by Theorem 3, = E(x log a + x log b) by distributivity, = E(x log a) - E(x log b) by Theorem 2, = x - bx. by definition. :1 Theorem 5. Let c _b_e a Leg constant. Then D(xc) = cxc_1 if x > 0, c xc+1 Ix dx=m+0 if ”—1 and x>0. Proof. Since xc = E(c log x), we can use the chain rule. We have D(xc) = E(c log x)D(c log x) = E(c log x)c/x = x°(c/x) = cxc-l. The integration formula follows at once. 0 Theorem6. Mamaflm. a> um D(a.x) = ax log a, 'axdx-l—ax- +C 'faaél -083 l . The proof is left as an exercise For other differentiation and integration formulas involving logarithms and, _ exponentials, see 6.7 and 6.16 of Apostol. M.10 M concerning (w—mnfl logarithms. The logarithm function we have defined is sometimes called the "natural logarithm". A different version of the logarithm was once useful. It can be obtained as follows: Consider the function {(x) = 10K = E(x log 10). It is strictly increasing, since by the chain rule, f'(x) = E(x log 10) - log 10, which is positive. Furthermore, since it log 10 takes on all real values, E(x log 10) = {(x) takes on all positive values. The inverse of f is called the "common logarithm" or the "logarithm to the base 10", and denoted by log10 y. That is, if y > 0, we define 10310 y = x if and only if y = 10". This function was at one time useful for computational purposes, but it has long since fallen into oblivion. A similar remark applies to obtain logarithms to other bases. If b is any positive number with b ,t 1, one defines logb y = x if and only if y = bx. Remark n motivation Just as the sine and cosine functions arise most naturally as the fundamental solutions of the differential equation for simple harmonic motion, so the exponential function arises most naturally from consideration'of the important differential equation £’(x) = kf(x), called the "equation of population growth (or decay)" If k is, for instance, the difference of the birth and death rate (per thousand, say, of a population) in a given time period, then this is the equation for the actual population (in thousands), as a function of time. M.11 One checks at once that the function ekx satisfies this equation. More generally, e_ve_r1 solution of this equation can be expressed in terms of e19: Theorem Z. Suppose {(x) is defined for all x and satisfies the equation {’(x) = kf(x). Let {(0) = a. Then f(x) = aekx. Proof. Let us set 3(X) = f(DOM-10‘)- Then we compute g’(x) = r’(x)E(—kx) — Htx)E’(—kx) = 0, so 3 is constant. Since g(0) = f(0)-E(0) = a, so!) = a. . Multiplying by E(kx), we have {(x) = aE(kx), as desired. -u g Exercises i 1. If a is constant, show that in general D(a.x) at xax—l x+1 I ax dx 1‘ amn- [Anyone who makes the mistake, on a quiz, of thinking these are equalities gets clobberedl] ,and 1x ' 171' 2. Evaluate] 7r dx and] x dx. 0 0 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern