exponential function defined as limit of powers


It is possible to define the exponential functionDlmfDlmfMathworldPlanetmathPlanetmath and the natural logarithmMathworldPlanetmathPlanetmathPlanetmath in terms of a limit of powers. In this entry, we shall present these definitions after some background information and demonstrate the basic properties of these functionsMathworldPlanetmath from these definitions.

Two basic results which are needed to make this development possible are the following:

Theorem 1.

Let x be a real number and let n be an integer such that n>0 and n+x>0. Then

(n+xn)n<(n+1+xn+1)n+1.
Theorem 2.

Suppose that {sn}n=1 is a sequenceMathworldPlanetmath such that limnnsn=0. Then limn(1+sn)n=1,

For proofs, see the attachments. From them, we first conclude that a sequence converges.

Theorem 3.

Let x be any real number. Then the sequence

{(n+xn)n}n=1

is convergentMathworldPlanetmathPlanetmath.

The foregoing results show that the limit in the following definition converges, and hence defines a bona fide function.

Definition 1.

Let x be a real number. Then we define

exp(x)=limn(n+xn)n.

We may now derive some of the chief properties of this function. starting with the addition formulaPlanetmathPlanetmath.

Theorem 4.

For any two real numbers x and y, we have exp(x+y)=exp(x)exp(y).

Proof.

Since

n(n+x+y)(n+x)(n+y)=1-xy(n+x)(n+y)

and

limnn(n+x)(n+y)=0,

theorem 2 above implies that

limn(n(n+x+y)(n+x)(n+y))n=1.

Since it permissible to multiply convergent sequences termwise, we have

exp(x)exp(y) =limn(n+xn)n(n+yn)n(n(n+x+y)(n+x)(n+y))n
=limn(n+x+yn)n=exp(x+y)

Theorem 5.

The function exp is strictly increasing.

Proof.

Suppose that x is a strictly positive real number. By theorem 1 and the definition of the exponentialMathworldPlanetmathPlanetmath as a limit, we have 1+x<exp(x), so we conclude that 0<x implies 1<exp(x).

Now, suppose that x and y are two real numbers with x>y. Since x-y>0, we have exp(x-y)>1. Using theorem 4, we have exp(x)=exp(x-y)exp(y)>exp(y), so the function is strictly increasing. ∎

Theorem 6.

The function exp is continuousMathworldPlanetmathPlanetmath.

Proof.

Suppose that 0<x<1. By theorem 1 and the definition of the exponential as a limit, we have 1-x<exp(-x) and 1+x<exp(x). By theorem 4, exp(x)exp(-x)=exp(0)=0. Hence, we have the bounds 1+x<exp(x)<1/(1-x) and 1-x<exp(-x)<1/(1+x). From the former bound, we conclude that limx0-exp(x)=1 and, from the latter, that limx0+exp(x)=1, so limx0exp(x)=1.

Suppose that y is any real number. By theorem 4, exp(x+y)=exp(x)exp(y). Hence, limx0exp(x+y)=exp(y)limx0exp(x)=exp(y). In other words, for all real y, we have limxyexp(x)=exp(y), so the exponential function is continuous. ∎

Theorem 7.

The function exp is one-to-one and maps onto the positive real axis.

Proof.

The one-to-one property follows readily from monotonicity — if exp(x)=exp(y), then we must have x=y, because otherwise, either x<y or x>y, which would imply exp(x)<exp(y) or exp(x)>exp(y), respectively. Next, suppose that x is a real number greater than 1. By theorem 1 and the definition of the exponential as a limit, we have 1+x<exp(x). Thus, 1<x<exp(x); since exp is continuous, the intermediate value theorem asserts that there must exist a real number y between 0 and x such that exp(y)=x. If, instead, 0<x<1, then 1/x>1 so we have a real number y such that exp(y)=1/x. By theorem 4, we then have exp(-y)=x. So, given any real number x>0, there exists a real number y such that exp(y)=x, hence the function maps onto the positive real axis. ∎

Theorem 8.

The function exp is convex.

Proof.

Since the function is already known to be continuous, it suffices to show that exp((x+y)/2)(exp(x)+exp(y))/2 for all real numbers x and y. Changing variables, this is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to showing that 2exp(a+b)exp(a)+exp(a+2b) for all real numbers a and b. By theorem 4, we have

exp(a+b) =exp(a)exp(b) (1)
exp(a+2b) =exp(a)exp(b)2. (2)

Using the inequalityMathworldPlanetmath 2x1+x2 with x=exp(b) and multiplying by exp(a), we conclude that 2exp(a+b)exp(a)+exp(a+2b), hence the exponential function is convex. ∎

Defining the constant e as exp(1), we find that the exponential function gives powers of this number.

Theorem 9.

For every real number x, we have exp(x)=ex.

Proof.

Applying an inductionMathworldPlanetmath argumentMathworldPlanetmathPlanetmath to theorem 4, it can be shown that exp(nx)=exp(x)n for every real number x and every integer n. Hence, given a rational number m/n, we have exp(m/n)n=exp(m)=exp(1)m=em. Thus, exp(m/n)=em/n so we see that exp(x)=ex when x is a rational number. By continuity, it follows that exp(x)=ex for every real number x. ∎

Title exponential function defined as limit of powers
Canonical name ExponentialFunctionDefinedAsLimitOfPowers
Date of creation 2013-03-22 17:01:35
Last modified on 2013-03-22 17:01:35
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 27
Author rspuzio (6075)
Entry type Definition
Classification msc 32A05
Related topic ExponentialFunction
Related topic ComplexExponentialFunction
Related topic ExponentialFunctionNeverVanishes