The exponential

In document Basic Analysis: Introduction to Real Analysis (Page 188-192)

The Riemann Integral

5.4 The logarithm and the exponential

5.4.2 The exponential

Just as with the logarithm we define the exponential via a list of properties. Proposition 5.4.2. There exists a unique function E: R(0,∞)such that

(i) E(0) =1.

(ii) E is differentiable and E0(x) =E(x). (iii) E is strictly increasing, bijective, and

lim

x→−∞E(x) =0, and xlim→∞E(x) =∞. (iv) E(x+y) =E(x)E(y)for all x,yR.

(v) If qQ, then E(qx) =E(x)q.

Proof. Again, we prove existence of such a function by defining a candidate, and prove that it satisfies all the properties. TheLdefined above is invertible. LetE be the inverse function ofL. Property is immediate.

Property follows via the inverse function theorem, in particular : Lsatisfies all the hypotheses of the lemma, and hence

E0(x) = 1

L0 E(x) =E(x).

Let us look at property . The functionE is strictly increasing sinceE0(x) =E(x)>0. AsE is the inverse ofL, it must also be bijective. To find the limits, we use thatE is strictly increasing and onto(0,∞). For everyM>0, there is anx0such thatE(x0) =MandE(x)≥Mfor allx≥x0. Similarly, for everyε >0, there is anx0such thatE(x0) =ε andE(x)<ε for allx<x0. Therefore,

lim

n→−∞E(x) =0, and nlim→∞E(x) =∞.

To prove property we use the corresponding property for the logarithm. Takex,yR. AsL

is bijective, findaandbsuch thatx=L(a)andy=L(b). Then E(x+y) =E L(a) +L(b)

=E L(ab)

=ab=E(x)E(y).

Property also follows from the corresponding property ofL. GivenxR, letabe such that

x=L(a)and

E(qx) =E qL(a)E L(aq

)

=aq=E(x)q.

Finally, uniqueness follows from and . LetE andF be two functions satisfying and . d

dx

F(x)E(x)=F0(x)E(x)E0(x)F(x) =F(x)E(x)E(x)F(x) =0.

Therefore, by ,F(x)E(−x) =F(0)E(−0) =1 for all x∈R. Next, 1=E(0) =

E(xx) =E(x)E(x). Then

0=1−1=F(x)E(−x)−E(x)E(−x) = F(x)−E(x)

Finally,E(−x)6=0 for allx∈R. SoF(x)−E(x) =0 for allx, and we are done.

Having provedE is unique, we define theexponentialfunction as exp(x):=E(x).

IfyQandx>0, then

xy=exp ln(xy)

=exp yln(x) .

We can now make sense of exponentiationxyfor arbitraryy

R; ifx>0 andyis irrational, define

xy:=exp yln(x) .

As exp is continuous thenxy is a continuous function ofy. Therefore, we would obtain the same result had we taken a sequence of rational numbers{yn}approachingyand definedxy=limxyn.

Define the numbere, sometimes calledEuler’s numberor thebase of the natural logarithm, as e:=exp(1).

Let us justify the notationexfor exp(x):

ex=exp xln(e)

=exp(x).

Let us extend properties of logarithm and exponential to irrational powers. The proof is immediate.

Proposition 5.4.3. Let x,yR.

(i) exp(xy) = exp(x)y

.

(ii) If x>0thenln(xy) =yln(x).

Remark 5.4.4. There are other equivalent ways to define the exponential and the logarithm. A common way is to defineE as the solution to the differential equationE0(x) =E(x),E(0) =1. See , for a sketch of that approach. Yet another approach is to define the exponential function by power series, see .

Remark5.4.5. We have proved the uniqueness of the functions LandE from just the properties L(1) =0,L0(x) =1/x and the equivalent condition for the exponential E0(x) =E(x), E(0) =1. Existence in fact also follows from just these properties. Alternatively, uniqueness also follows from the laws of exponents, see the exercises.

Eis a function into(0,)after all. However,E(

−x)6=0 also follows fromE(x)E(x) =1. Therefore, we can prove uniqueness ofEgiven and , even for functionsE:R→R.

5.4.3

Exercises

Exercise5.4.1: Let y be any real number and b>0. Define f: (0,∞)Rand g:R→Ras, f(x):=xy

and g(x):=bx. Show that f and g are differentiable and find their derivative.

Exercise5.4.2: Let b>0, b6=1be given.

a) Show that for every y>0, there exists a unique number x such that y=bx. Define thelogarithm baseb,

logb: (0,∞)→R, bylogb(y):=x.

b) Show thatlogb(x) = lnln((xb)).

c) Prove that if c>0, c6=1, thenlogb(x) =logc(x)

logc(b).

d) Provelogb(xy) =logb(x) +logb(y), andlogb(xy) =ylogb(x).

Exercise5.4.3(requires ): Use to study the remainder term and show that for all xR

ex= ∞

n=0 xn n!.

Hint: Do not differentiate the series term by term (unless you would prove that it works).

Exercise5.4.4: Use the geometric sum formula to show (for t6=1)

1−t+t2− ···+ (1)ntn= 1

1+t−

(1)n+1tn+1

1+t .

Using this fact show

ln(1+x) = ∞

n=1 (1)n+1xn n

for all x∈(1,1](note that x=1is included). Finally, find the limit of the alternating harmonic series

n=1 (1)n+1 n =1−1/2+1/3−1/4+··· Exercise5.4.5: Show ex=nlim →∞ 1+x n n .

Hint: Take the logarithm. Note: The expression 1+xnn

arises in compound interest calculations. It is the amount of money in a bank account after 1 year if 1 dollar was deposited initially at interest x and the interest was compounded n times during the year. The exponential exis the result of continuous compounding.

Exercise5.4.6:

a) Prove that for n∈Nwe have

n

k=2 1 k ≤ln(n)≤ n−1

k=1 1 k. b) Prove that the limit

γ:= lim n→∞ n

k=1 1 k−ln(n) !

exists. This constant is known as theEuler–Mascheroni constant . It is not known if this constant is rational or not. It is approximatelyγ≈0.5772.

Named for the Swiss mathematician (1707–1783) and the Italian mathematician (1750–1800).

Exercise5.4.7: Show

lim

x→∞

ln(x)

x =0.

Exercise5.4.8: Show that exisconvex, in other words, show that if a≤x≤b then exea b−x

b−a+eb xb−−aa.

Exercise5.4.9: Using the logarithm find

lim

n→∞n 1/n.

Exercise5.4.10: Show that E(x) =exis the unique continuous function such that E(x+y) =E(x)E(y)and

E(1) =e. Similarly, prove that L(x) =ln(x)is the unique continuous function defined on positive x such that

L(xy) =L(x) +L(y)and L(e) =1.

Exercise 5.4.11(requires ): Since(ex)0 =ex, it is easy to see that ex is infinitely differentiable (has

derivatives of all orders). Define the function f:R→R.

f(x):=

(

e−1/x if x>0,

0 if x≤0.

a) Prove that for any m∈N,

lim

x→0+

e−1/x

xm =0.

b) Prove that f is infinitely differentiable.

c) Compute the Taylor series for f at the origin, that is,

k=0 f(k)(0)

k! xk.

In document Basic Analysis: Introduction to Real Analysis (Page 188-192)