The Exponential Function

Transcription

1 The Expoetial Fuctio A. D. Adrew School of Mathematics Georgia Istitute of Techology September Itroductio. I these otes we 1. Review the basics of the expoetial fuctio a where a > 0, a atural umber, see how to exted it to ratioal, the irratioal, ad fially egative expoets.. Show how to calculate its derivative. 3. Itroduce the importat umber e, the so-called base of the atural logarithm, as the limit of a sequece. 4. See how this limit (ad hece the expoetial fuctio) arises i the cotext of compoud iterest. We will use the otio of covergece of a sequece of real umbers, ad the Least Upper Boud Axiom for the real umbers. If you are ot familiar with these topics, you should look them up i your calculus text. 1. The expoetial fuctio. Whe a > 0, ad = 1,,3,L is a atural umber, the expoetial a = a 14 a L 43 a times is simply the product of a with itself times. For a, b > 0, this fuctio satisfies the usual rules

2 The Expoetial Fuctio (1) a. a + m = a a m, b. a ( ) m = a m, c. a b = ( ab), d. If a 1, the a = a m if ad oly if = m, e. If a > 1, the a is a icreasig fuctio of, f. If a < 1, the a is a decreasig fuctio of. O the other had, if we regard as fixed, ad a > 0 as the variable, a is a icreasig fuctio of a. Usig this observatio ad the Least Upper Boud Axiom 1, we see that for each atural umber ad each real umber a > 0, there is a uique positive real umber b such that b = a. We call b the -th root of a, ad deote it by a. Specifically, a = Least Upper Boudof theset{ x :x a}. Next for a ratioal expoet, that is, oe which is the ratio of two o-egative atural umbers as a p q q = ( a) p q = a p, ad we leave it as a exercise for the reader to covice itself that the rules (1) still hold whe the expoets are allowed to be ratioal umbers. For real umbers r which are ot ratioal, such as or π, we agai look to the icreasig ature of the expoetial fuctio (1e above) ad the Least Upper Boud Axiom ad defie, for a > 1 a r = Least UpperBoud of theset { a s : s<r, s ratioal. }. 1 Recall that the Least Upper Boud Axiom states that every o-empty set of real umbers which has a upper boud has a least upper boud. As a example, the umbers 5, π, ad Avogadro's costat , are examples of upper bouds for the set A = { x :x = 1 1, = 1,,3,... } = { 1, 3, 3 4,...}. The least upper boud for this set is 1, sice o smaller umber is a upper boud, ad every other upper boud is at least as large as 1. Please see your calculus text for further explaatio.

3 The Expoetial Fuctio 3 We leave it to the reader to figure out how to defie a r the expoet r is egative ( r < 0), we defie for 0 < a < 1, ad fially whe a r = 1 a r. Notice that sice r < 0, the right had side of this equatio is already defied. Agai, please covice yourself that the rules i (1) are valid whe the expoets are ay real umbers. Sketched below are the graphs of expoetial fuctios with bases a = 1, ad 3. Which is which?. The derivative of the expoetial fuctio. From the above picture it is ituitively clear that expoetial fuctios have well defied taget lies, ad hece, the derivative of the expoetial--the slope of this taget lie-- exists. Let's try to calculate it.

4 The Expoetial Fuctio 4 d dx ax a x + h a x = lim h 0 h a x a h a x = lim h 0 h a h 1 = a x lim h 0 h a h a 0 = a x lim. h 0 h a h 1 Now lim is just a costat. I fact, it's the derivative of a x evaluated at a = 0. h 0 h That is, the derivative of the expoetial fuctio is d dx ax = k a a x, a costat multiple of itself, ad that costat (which depeds o the base a) is the slope at the poit x = 0. From the graph above, that slope at 0 is positive ad large for some a, ad egative for other a. Perhaps the most atural is that value of a for which the derivative at 0 is 1. This is the so called base of the atural logarithm, deoted by e. If you draw the above graph carefully, you'll see that the derivative of x at 0 is less tha 1, ad the derivative of 3 x at 0 exceeds 1. Thus < e < 3. Exercises. 1. Carefully sketch (by had or by computer) the graphs of f (x) = x ad g(x) = 3 x, usig equal scales o the axes. Carefully sketch the taget lies at x = 0, ad covice yourself that f ' (0) < 1 < g' (0). Deduce that < e < 3.. Sketch the graphs ad calculate the first ad secod derivatives. x 3x a. y = e cos(4 x) b. y = e si( x) c. sih( x) x x e e = (the hyperbolic sie) d. x x e + e cosh( x) = (the hyperbolic cosie)

5 The Expoetial Fuctio 5 3. The umber e. As outlied above, we defie e to be that umber for which e h 1 lim h 0 h = 1, ad by exercise 1, it is plausible that there is such a positive umber, ad that it lies somewhere betwee ad 3. Usig the symbol to mea "approximately equal," we have that for values of h ear 0, e h 1 h 1, so so we expect that we may calculate e as e h 1 + h, e ( 1 + h) 1 h, e = lim 1 + h h 0 ( ) 1 h = lim ( ) the limit of a sequece. You might (icorrectly) thik that lim ( ) should be 1, sice the terms become close to 1, ad 1 to ay power is 1. You might (also icorrectly) thik that lim ( ) should be ifiite sice each term is strictly bigger that 1, ad powers icreasig without boud ted to ifiity. Remember, we saw i Exercise 1 that < e < 3, so these argumets must both be wrog. The poit is that the base is decreasig towards 1 at the same time as the expoet is icreasig to. These actios are ot idepedet, as assumed i the specious argumets above. Here is a proof that lim ( ) actually does exist. The value, of course, is the umber we call e. I our argumet, we will use Beroulli's iequality, which states that for a > 1, a 0, (1 + a) k > 1 + ka. Defie two sequeces ( 1 ) ( 1 + s = 1 +, t = 1+ ) 1. We will show that

6 The Expoetial Fuctio 6 a. the s form a strictly icreasig sequece, b. the t form a strictly decreasig sequece, c. s < t for each. { } ad { t } are bouded, mootoe sequeces, ad thus have Cosequetly s limits. Sice t = s ( ), their limits are the same -- that umber we call e, ad sice s < e < t we ca calculate s ad t ad thus approximate e to as may decimal places as we choose. To show that the s form a strictly icreasig sequece, we ote s +1 s = 1 ( ) + 1 ( ) = ( ) = +1 = +1 > +1 = ( ) ( ) ( ) (by Beroulli' s Iequality with a =-, ad k =+1) (+ 1) Thus s +1 > s ad the sequece is icreasig. We leave the proofs of statemets b ad c to the reader. Exercises. 3. Beroulli's iequality ca be proved by mathematical iductio. Use mathematical iductio (look it up i your calculus text if is ot familiar to you) to prove Beroulli's iequality. 4. Use the iequalities s < e < t to approximate e to oe decimal place. What's the smallest you ca use?

7 The Expoetial Fuctio 7 5. By chagig variables, show that for ay r, e r = lim ( ) 1 h that the limits lim 1 + h h 0 both equal to e. 4. Compoud Iterest. (h real) ad lim ( ). You may assume 1 + r ( ) ( a atural umber) are The basic formula for simple iterest is i = Pr t. That is, iterest equals pricipal times rate times time. For example, $P, ivested for t years at a iterest rate of r per cet per year (expressed as a decimal), yields iterest i. Thus $1,000 ivested at 4% for years yields iterest of (1000)(.04)() = $80, ad the ivestmet is the worth $1080. More commoly, iterest is compouded. If, for example, iterest is compouded aually, oe begis year 1 with $P ad eds year 1 with $(P +P i) = P(1 + i). This pricipal is ivested for all of the secod year, at the ed of which the ivestmet's value is ( P + Pi) + ( P + Pi)i = P( 1 + i) I geeral, the value of the ivestmet after t years is P( 1+i) t. If our $1,000 is ivested at 4% for two years compouded aually, the ivestmet grows to a value of $ , so compoudig is clearly i the ivestor's favor. If iterest is compouded quarterly ( four times per year), the the aual iterest rate r is a quarterly rate of r 4, ad the aalysis above shows that after k quarters, the value of the ivestmet is P 1 + r 4 Thus $1,000 ivested at 4% compouded quarterly for two years (8 quarters) grows to a value of k (4 )() = More geerally, if iterest is compouded m times per year, the value of the ivestmet after t years is P 1 + r m m t,

8 The Expoetial Fuctio 8 ad i what is called cotiuous compoudig, ( the limit as m teds to ), the value of the ivestmet after t years is Pe r t by the result of exercise 4. Exercises. 6. O the same set of axes, plot the values of a ivestmet of $1, ivested at 6% per year compouded aually, quarterly, mothly, daily, ad cotiuously, as a fuctio of time t measured i years. 7. At age 0, you wish to ivest $P, to be compouded cotiuously at 6% per year, i order to be able to pay for your childres' educatio, begiig whe you are 50. Make reasoable assumptios as to the cost of educatio (o Hope scholarship!), umber of childre (positive), ad calculate P. Do ot take icome tax ito accout. 8. Suppose istead that at age 0, you begi ivestig $Q at the begiig of each year util you are 50. Usig the same cost of educatio, umber of childre, ad iterest rate as i problem 6, calculate the amout Q you must set aside i order to pay for your childres' educatio. Agai, igore taxes. (Hit: Thik of doig problem 6 at ages 0, 1,,...) Refereces. Johso, Kiokemeister, Calculus with aalytic geometry, secod editio, Ally ad Baco, Ic., Salas, Hille, Etge, Calculus: oe ad several variables, ith editio, Joh Wiley ad Soes, Ic., 003.