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Chapter 13

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

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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 ...
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