(a 1 m. a n m = < a 1/N n

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1 Notes on a an log a Mat 9 Fall 2004 Here is an approac to te eponential an logaritmic functions wic avois any use of integral calculus We use witout proof te eistence of certain limits an assume tat certain functions on te rational numbers can be etene to continuous functions on te reals All of tis can be justifie, but we o not o so ere Let a be a positive real number We want to efine a For n a positive integer, a n is a multiplie by itself n times Similarly, a n is a multiplie by itself n times Every positive number as a unique positive m-t root, so we can efine for m a positive integer a n m = a m n Having efine a for a rational number, we efine a for all real by coosing a sequence of rationals converging to, etc Tis process leas to a well efine function a wic is a continuous function from te wole real line to te positive reals Proposition a +y = a a y an a y = a /a y Moreover, a 0 = an a = a Proof Te first two assertions follow by first proving tem for an y rational an ten using continuity To sow a 0 =, set an y bot equal to in te secon ientity Tat a = a follows from te efinition Lemma 2 Assume a > Ten a is a strictly increasing function Proof Suppose < y an tat an y are rational By aing a positive integer to bot sies, we can assume tat an y are positive Write bot over a common enominator N Tus, = m/n an y = n/n Since < y, we ave m < n Now, a > implies a /N > Tus, a = a /N m < a /N n = a y Having establise te result for rational numbers, te general result follows by taking limits Definition For a > 0, te limit a lim eists It is calle lna, te natural logaritm of a Proposition 3 Te erivative of a eists, an we ave a = a lna Proof Tis follows easily from te efinitions a + a a = lim = a lim 0 a = a lna

2 Remark: note te above proposition is similar to ow we ifferentiate trigonometric functions To fin te erivative of sin or cos at any point, we neee to compute sin sin 0 cos cos 0 two limits, lim an lim ; te erivative of sine an cosine at any point followe by te angle aition formulas Here, te analogue is a + = a a Corollary If a > ten lna > 0 For a =, ln = 0 If 0 < a <, ten lna < 0 Proof Te secon assertion is clear since = for all an so its erivative is zero To prove te first assertion, note tat if lna < 0 te by te Proposition, a is ecreasing Tis contraicts Lemma 2 Tus, lna 0 if a > However, it can t be 0, since ten te erivative of a woul be ientically zero by Proposition 3, an a woul be a constant Tis again contraicts Lemma 2 Tus, lna > 0 Finally, if a <, we ave a = a is ecreasing, so, by repeating te reasoning above, we fin lna < 0 Remark: We o not consier lna for a 0 Proposition 4 Assume a, b > 0 Ten lnab = lna + lnb Proof By efinition, As we ave Terefore [ a b lnab = lim ab lnab = lim ab = ab b + b, ab = a b + b ] + b a = lim As 0, note b, an we terefore fin tat Ten, lnab = lna + lnb logab = lna + lnb b b + lim Proof We ave alreay prove te first assertion, but ere is a secon proof using Proposition 4 ln = ln = ln + ln, wic implies ln = 0 Now, 0 = ln = lna a = lna + lna Te secon assertion follows Lemma 5 ln is a strictly increasing function Proof Suppose 0 < < y Ten, < y/ an lny/ > 0 by te Corollary to Proposition 3 Tus lny ln > 0, or ln < lny, wic proves te assertion 2

3 Proposition 6 For all positive a, we ave lna = lna Proof Let fw = lna Using Proposition 4, one easily cecks tat fw + w 2 = fw + fw 2 By simple algebra one euces tat fr = rf = r lna for all rational numbers r Since fw is continuous, te Proposition follows for all real numbers Corollary As, ln Also, as 0, ln Proof Since ln is increasing we only ave to prove tat it takes on arbitrarily large values as gets bigger an bigger Consier te sequence 2, 2 2, 2 3,, 2 n, We ave ln2 n = n ln2 as n Similarly, on te sequence 2, 2 2, 2 3,, 2 n, we ave Tis completes te proof Definition Te following limit eists ln2 n = n ln2 as n lim + 0 We call tis limit e, Euler s constant; e is approimately epression for e is given by e = lim + n n n Proposition 7 Te function ln is ifferentiable We ave lne ln = Proof As + = +, using Proposition 4 we ave ln + ln = ln + ln = ln + ln Terefore we ave ln + ln = ln + Using te efinition of te erivative, we see tat ln + ln ln = lim + = ln + = lim 0 Note tat anoter ln = ln + ln + Letting =, as 0 we see 0 By Proposition 6, lim 0 ln + = lim ln +, 0 an by te above efinition tis limit is just lne Combining te pieces gives te erivative of ln is lne, as claime 3

4 Proposition 8 lne = Proof By Proposition 6, we ave lne = lne Differentiate bot sies using wat we ave proven an, of course, te cain rule Remember tat e is a constant, so lne is just a number it as no epenence Tus, te erivative wit respect to of lne is zero, an te erivative wit respect to of lne is tus lne We use te cain rule to ifferentiate lne Let f = ln an let g = e Ten lne = fg, so by te cain rule its erivative is f g g We get te erivative of f from Proposition 7, an te erivative of g from Proposition 3 Substituting gives [lne ] = lne e e lne We ave sown tat tis erivative is also equal to lne; terefore we fin tat [lne] 2 = lne Since lne 0, we must ave lne = Corollary ln = an e = e Definition We efine te logaritm function to te base a by te following formula log a = ln lna Tis function as all te properties one woul epect We list tem Te proofs are very easy an are left to te reaer log a = log a 2 log a y = log a + log a y an log a y 3 log a = 0 an log a a = 4 log a a = 5 a log a = 6 [log a] = = log a log a y lna 7 If a >, ten log a is strictly increasing an its grap is everywere concave own It goes to as an to minus as 0 4

5 Finally, we note tat ln = log e, so tat we ave e ln = an lne = Wat tis means is tat if log a = y ten = a y In oter wors, log a is te number of powers of a we nee to get For eample, consier a log a : we raise a to te number of powers of a we nee to get ; tus a log a = Finally, note tat in general log a + y log a + log a y; te logaritm of a sum is not te sum of te logaritms 5