The Exponential Function as a Limit

Transcription

1 Applied Mathematical Scieces, Vol. 6, 2012, o. 91, The Expoetial Fuctio as a Limit Alvaro H. Salas Departmet of Mathematics Uiversidad de Caldas, Maizales, Colombia Uiversidad Nacioal de Colombia, Maizales FIZMAKO Research Group asalash2002@yahoo.com Abstract I this paper we defie the expoetial fuctio of base e ad we establish its basic properties.we also defie the logarithmic fuctio of base e ad we prove its cotiuity. Keywords: umber e, it of sequece of fuctios, expoetial fuctio, logarithmic fuctio 1 Itroductio Let N = { 1, 2, 3,...} be the set of atural umbers ad let R be the set of real umbers. Suppose that { f x) } =1 is a sequece of fuctios defied o E R. We say that this sequece coverges to the fuctio fx) oe if This meas that f x) =fx) for ay x E. ε>0 x E N = Nε, x) N N : >N f x) fx) <ε 1.1) I this case we write f x) E fx) ). Suppose that x R. Let us cosider the umbers m 0 = m 0 x) ad 0 = 0 x) defied as follows : m 0 = m 0 x) ={k N k>x} ad 0 = 0 x) ={k N k> x} 1.2) Thus, m 0 = 1 ad 0 =[ x] +1if x 0 ad m 0 =[x] + 1 ad 0 =1if x 0[x] is the iteger part of x). It is clear that 1 + x > 0 for ay 0

2 4520 A. H. Salas ad 1 x > 0 for ay m 0. We defie the two sequeces { f x) } =1 ad { g x) } =1 as follows : f x) =0 if < 0 ad f x) = 1+ ) x if ) Moreover, g x) =0 if <m 0 ad g x) = 1 x ) if m0. 1.4) Lemma 1. Let x R ad cosider the sequeces defied by 1.3) ad 1.4). a. The sequece { f x) } =1 is icreasig for 0, that is, f x) f +1 x) for ay 0. I particular it is icreasig for x 0 sice 0 =1. b. The sequece { g x) } =1 is decreasig for m 0, that is, g x) g +1 x) for ay m 0. I particular it is icreasig for x 0 sice m 0 =1. c. 0 g x) f x) x2 g k 0 x) for ay k 0 = maxm 0, 0 ). d. There exist the its f x) = sup{ f x) N } ad g x) = f x) =L. Moreover, f 0 x) L g m0 x) e. If h < 1 the 1+h 1+ ) h 1 ) h 1 h) 1 for all 1 1.5) Proof. a. Let 0. From the AGM iequality a 1 + a a with +1 a 1 a 2 a +1 a i > 0, i =1, 2,..., + 1) 1.6) a 1 =1, a 2 = a 3 = = a +1 =1+ x > 0 we obtai ad the x 1+ x = ) +1 f +1 x) = x ) 1+ x ) x = f x). +1 ) This iequality is strict uless x =0.

3 The expoetial fuctio as a it 4521 b. Let m 0. From the AGM iequality 1.6) with a 1 =1, a 2 = a 3 = = a +1 =1 x > 0. it follows that ad the x 1 x = ) x ), 1 x ) +1 1 x > 0, +1 ) which implies g +1 x) = 1 x ) +1) 1 x = g x). +1 ) This iequality is strict uless x =0. c. Let k 0 = maxm 0, 0 ). We have g x) f x) =g x) 1 f ) x) = g x)1 q ), 1.7) g x) where q =1 x2 2. Observe that k 0 > x from where 0 <q 1 ad the q 1 ad 1 q 0. It is clear from 1.7) that g x) f x) 0 for k 0. O the other had, by virtue of 1.7), Thus, 0 g x) f x) = g x)1 q)1+q + + q 1 ) g k0 x) x ) 2 = g k0 x) x2 = x2 g 2 k 0 x). 0 g x) f x) x2 g k 0 x) for k ) From the last iequality we see that give ε>0if we choose a atural umber N subject to N k 0 ad N>x 2 g k0 x)/ε the We have proved that g x) f x) = g x) f x) <ε for all >N. g x) f x)) = )

4 4522 A. H. Salas d. Let k 0 = maxm 0, 0 ) m 0. By virtue of b ad c, g x) g k0 x) ad f x) g x) g k0 x) for all k 0, which proves that the sequece { f x) } =1 is bouded from above for each x R ad the f x) =L, where L = sup{ f x) N } = sup{ f x) 0 }. O the other had, from 1.9), g x) = g x) f x)) + f x)) = L. e. Observe that m 0 = 0 = 1 sice h < 1. From a, b ad c we obtai 1+h = f 1 h) f h) g h) g 1 h) =1 h) 1 for ay k 0 =1. 2 The expoetial fuctio ad its properties I previous sectio we established the existece of the its 1+ x = 1 ) x ) for each x R. This allows us to defie a fuctio exp : R 0, ) as follows : expx) = 1+ x ) = 1 x ),x R. 2.1) Is obvious that exp0) = 1. The value exp1) is special ad it is deoted by e : e = ) We well call fuctio defied by 2.1) the expoetial fuctio of base e. This fuctio is also deoted by e x. 2.1 Properties of the expoetial fuctio I this sectio we establish the mai properties of the expoetial fuctio startig from 2.1). Property 1. Let x R. i. If x> 1 the expx) > 1+x. I particular, expx) > 1 for x>0.

5 The expoetial fuctio as a it 4523 ii. If x<1 the expx) 1. I particular, expx) < 1ifx<0. 1 x Proof. i. Sice x> 1 we have 0 =[ x] + 1 = 1. By virtue of lemma 1, parts a ad d, expx) 1+ x ) 2 > 1+ x ) 1 =1+x. 2 1 ii. If x<1the m 0 =[x] + 1 = 1. I view of lemma 1 parts b, c ad d, f x) g x) g 1 x) for all k 0 = maxm 0, 0 ) ad lettig we obtai expx) = f x) g 1 x) = 1 x ) 1 1 = 1 1 x. Property 2. Multiplicative property ) expx + y) = expx) expy) = expy) expx) for ay x, y R. 2.2) I particular, exp x) = expx)) 1 = 1 expx) for all x R. 2.3) Proof. Let us cosider the sequeces f x) = 1+ ) x, f y) = 1+ ) y ad f x + y) = 1+ x + y ), where k 0 > x + y. By lemma 1, part d, f x) = expx), f y) = expy) ad f x + y) = expx + y). Sice h) def xy = 0 ),we may choose N large eough so + x + y that h) < 1 for N.We obtai f x)f y) f x + y) = 1+ xy + x + y) ) = 1+ h) I view of lemma 1, part e, from 2.4) it is clear that ) for N. 2.4) 1+h) f x)f y) f x + y) 1 h)) 1 2.5)

6 4524 A. H. Salas Takig ito accout that 1 + h)) = 1 h)) 1 = 1 from 2.5) we f x)f y) obtai =1, from where f x + y) expx) expy) expx + y) = f x) f y) f x + y) = f x)f y) f x + y) =1. We have proved that expx) expy) = expx + y). Property 3. Give t, x R, ift<x, the expt) < expx), that is, the expoetial fuctio is strictly icreasig o R. Proof.Ifx>tthe x t>0 ad makig use of Property 1, expx t) > 1. We have expx) = expx t)+t) = expx t) expt) > 1 expt) = expt). Property 4. If x>0 the 0 < expx) 1 x expx). Proof. Let N. We have 0 < 1+ x ) 1 = 1+ x ) 1 1+ x ) x ) ) < x 1+ x ) + 1+ x ) x ) ) = x 1+ x ) = x 1+ x <xexpx). ) Thus, 0 < 1+ x ) 1 <xexpx) for ay N. Lettig i the last iequality gives 0 < expx) 1 x expx). 2.6) Property 5. The expoetial fuctio is cotiuous o R, i.e, for a give real umber a ad ay ε>0 we may fid δ = δε, a) > 0 such that if x a <δ the expx) expa) <ε. Proof. Let us first show that expt) 1 3 t for t < ) Ideed, this iequality is obvious if t = 0. Let t 0.If0<t<1 the expt) < exp1) = e<3. Cosequetly, i view of Property 4, 0 < expt) 1 < 3t.

7 The expoetial fuctio as a it 4525 Now, let 1 <t<0. From oe had, by Property 1, expt) < 1.O the other had, 0 < t <1 ad the 0 < exp t) 1 < 3 t) =3 t, from where expt) 1 = expt)1 exp t)) = expt)exp t) 1) < 3 expt) t < 3 t. We have established 2.7). Let a R ad cosider values of x subject to x a < 1. Settig t = x a i 2.7) we obtai expx a) 1 < 3 x a. Multiplyig this iequality by expa) ad makig use of Property 2 we obtai expx) expa) < 3 expa) x a for ay x R such that x a < ) From coditio 2.8) it is clear that choosig δ such that ) 1 0 <δ<mi 2, ε, 3 expa) the expx) expa) for all x R such that x a <δ. This meas that expx) = expa). x a 3 The logarithmic fuctio I view of Property 5, the expoetial fuctio is strictly icreasig o R. I view of Property 1, part i, expx) > 1+x>xfor x 0. O the other had, if x<0 the 2.2) gives expx) exp x) = expx x) = exp0) = 1 ad the expx) > 0. This says that the expoetial fuctio exp : R 0, ) is oe to oe ad it admits iverse We will deote it by log ad we will call it logarithmic fuctio of base e : log : 0, ) R. Let y 0, ). There exists x R, uiquely defied, such that expx) =y. Ideed, choose b>0 subject to b>y 1. By Property 1, part i, expb) > 1+b>y. O the other had, let a be ay egative umber such that a<1 1/y. By Property 1, part ii, expa) 1 1 a <y. We have proved that for ay y 0, ) we may fid two real umbers a ad b such that a<bad expa) <y<expb). Let us cosider the fuctio expx) o the iterval [a, b]. Sice this fuctio is cotiuous o [a, b] Property 1 ), it takes all values betwee expa) ad expb). This allows us to choose x o [a, b] for which y = expx). This umber x is uique sice the expoetial

8 4526 A. H. Salas fuctio is oe to oe. We have proved that fuctio exp : R 0, ) is oe to oe ad oto. Thus, logy) =x y = expx). It is clear that explogy)) = y y>0 ad logexpx)) = x x R. Theorem. The logarithmic fuctio log : 0, ) R is cotiuous o 0, ). Proof : It is easy to see that give b > 0 ad ε > 0if y b < δ = mi b1 exp ε)),bexpε) 1)) the logy) logb) <ε.this meas that y b logy) = logb) for ay b>0. 4 Colusios We defied two of the most importat fuctios i mathematics: the expoetial ad logarithmic fuctios. This allows to defie the expoetial fuctio of base a as s a x = expx loga)),x R,a > 0 ad a 1. From this we may establish the laws of expoets : a. a x a y = a x+y ; b. ax a = y ax y ; c. a x b x =ab) x ; d. ax a ) x b = ; e. a x ) y = a xy. x b Fially, we may defie the logarithmic fuctio as a it as follows logx) = x 1) = 1 x 1/ ) for x>0. Startig from this defiitio we may defie the expoetial fuctio as the iverse of logarithmic fuctio. Refereces [1] Hardy G. H., A Course of Pure mathematics, Nith Editio, Uiversity of Cambridge 1945). [2]Mitriovic D.S., Elemetary Iequalities, Belgrade-Yugoslavia, P. Noordhoff Ltd - Groige - The Netherlads 1964) [3] Polya G. & Szego G., Aufgabe ud lehrsatze aus der aalysis, Spriger Verlag, Berli 1964). [4] Rudi W., Priciples of Mathematical Aalysis, Third Editio, McGraw Hill 1976). Received: April, 2012