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Characterizations of the exponential function

In mathematics, the exponential function can be characterized in many ways. The following characterizations (definitions) are most common. This article discusses why each characterization makes sense, and why the characterizations are independent of and equivalent to each other. As a special case of these considerations, we will see that the three most common definitions given for the mathematical constant e are also equivalent to each other.

Contents

CharacterizationsEdit

The six most common definitions of the exponential function exp(x) = ex for real x are:

1. Define ex by the limit
 
2. Define ex as the value of the infinite series
 
(Here n! denotes the factorial of n. One proof that e is irrational uses this representation.)
3. Define ex to be the unique number y > 0 such that
 
This is as the inverse of the natural logarithm function, which is defined by this integral.
4. Define ex to be the unique solution to the initial value problem
 
(Here, y′ denotes the derivative of y.)
5. The exponential function f(x) = ex is the unique Lebesgue-measurable function with f(1) = e that satisfies
 
(Hewitt and Stromberg, 1965, exercise 18.46). Alternatively, it is the unique anywhere-continuous function with these properties (Rudin, 1976, chapter 8, exercise 6). The term "anywhere-continuous" means that there exists at least a single point   at which   is continuous. As shown below, if   for all   and   and   is continuous at any single point   then   is necessarily continuous everywhere.
(As a counterexample, if one does not assume continuity or measurability, it is possible to prove the existence of an everywhere-discontinuous, non-measurable function with this property by using a Hamel basis for the real numbers over the rationals, as described in Hewitt and Stromberg.)
Because f(x) = ex is guaranteed for rational x by the above properties (see below), one could also use monotonicity or other properties to enforce the choice of ex for irrational x,[citation needed] but such alternatives appear to be uncommon.
One could also replace the conditions that   and that   be Lebesgue-measurable or anywhere-continuous with the single condition that  . This condition, along with the condition   easily implies both conditions in characterization 4. Indeed, one gets the initial condition   by dividing both sides of the equation
 
by  , and the condition that   follows from the condition that   and the definition of the derivative as follows:
 
6. Let e be the unique real number satisfying
 
This limit can be shown to exist. This definition is particularly suited to computing the derivative of the exponential function. Then define ex to be the exponential function with this base.

Larger domainsEdit

One way of defining the exponential function for domains larger than the domain of real numbers is to first define it for the domain of real numbers using one of the above characterizations and then extend it to larger domains in a way which would work for any analytic function.

It is also possible to use the characterisations directly for the larger domain, though some problems may arise. (1), (2), and (4) all make sense for arbitrary Banach algebras. (3) presents a problem for complex numbers, because there are non-equivalent paths along which one could integrate, and (5) is not sufficient. For example, the function f defined (for x and y real) as

 

satisfies the conditions in (5) without being the exponential function of x + iy. To make (5) sufficient for the domain of complex numbers, one may either stipulate that there exists a point at which f is a conformal map or else stipulate that

 

In particular, the alternate condition in (5) that   is sufficient since it implicitly stipulates that f be conformal.

Proof that each characterization makes senseEdit

Some of these definitions require justification to demonstrate that they are well-defined. For example, when the value of the function is defined as the result of a limiting process (i.e. an infinite sequence or series), it must be demonstrated that such a limit always exists.

Characterization 2Edit

Since

 

it follows from the ratio test that   converges for all x.

Characterization 3Edit

Since the integrand is an integrable function of t, the integral expression is well-defined. Now we must show that the function from   to   defined by

 

is a bijection. As   is positive for positive t, this function is monotone increasing, hence one-to-one. If the two integrals

 

hold, then it is clearly onto as well. Indeed, these integrals do hold; they follow from the integral test and the divergence of the harmonic series.

Equivalence of the characterizationsEdit

The following proof demonstrates the equivalence of the first three characterizations given for e above. The proof consists of two parts. First, the equivalence of characterizations 1 and 2 is established, and then the equivalence of characterizations 1 and 3 is established.

Equivalence of characterizations 1 and 2Edit

The following argument is adapted from a proof in Rudin, theorem 3.31, p. 63–-5.

Let   be a fixed non-negative real number. Define

 

By the binomial theorem,

 

(using x ≥ 0 to obtain the final inequality) so that

 

where ex is in the sense of definition 2. Here, we must use limsups, because we don't yet know that tn actually converges. Now, for the other direction, note that by the above expression of tn, if 2 ≤ mn, we have

 

Fix m, and let n approach infinity. We get

 

(again, we must use liminf's because we don't yet know that tn converges). Now, take the above inequality, let m approach infinity, and put it together with the other inequality. This becomes

 

so that

 

We can then extend this equivalence to the negative real numbers by noting   and taking the limit as n goes to infinity.

The error term of this limit-expression is described by

 

where the polynomial's degree (in x) in the term with denominator nk is 2k.

Equivalence of characterizations 1 and 3Edit

Here, we define the natural logarithm function in terms of a definite integral as above. By the first part of fundamental theorem of calculus,

 

Besides,  

Now, let x be any fixed real number, and let

 

We will show that ln(y) = x, which implies that y = ex, where ex is in the sense of definition 3. We have

 

Here, we have used the continuity of ln(y), which follows from the continuity of 1/t:

 

Here, we have used the result lnan = nlna. This result can be established for n a natural number by induction, or using integration by substitution. (The extension to real powers must wait until ln and exp have been established as inverses of each other, so that ab can be defined for real b as eb lna.)

 
 
 
 

Equivalence of characterizations 2 and 4Edit

Let n be a non-negative integer. In the sense of definition 4 and by induction,  .

Therefore  

Using Taylor series,   This shows that definition 4 implies definition 2.

In the sense of definition 2,

 

Besides,   This shows that definition 2 implies definition 4.

Equivalence of characterizations 1 and 5Edit

The following proof is a simplified version of the one in Hewitt and Stromberg, exercise 18.46. First, one proves that measurability (or here, Lebesgue-integrability) implies continuity for a non-zero function   satisfying  , and then one proves that continuity implies   for some k, and finally   implies k=1.

First, we prove a few elementary properties from   satisfying   and the assumption that   is not identically zero:

  • If   is nonzero anywhere (say at x=y), then it is non-zero everywhere. Proof:   implies  .
  •  . Proof:   and   is non-zero.
  •  . Proof:  .
  • If   is continuous anywhere (say at x = y), then it is continuous everywhere. Proof:   as   by continuity at y.

The second and third properties mean that it is sufficient to prove   for positive x.

If   is a Lebesgue-integrable function, then we can define

 

It then follows that

 

Since   is nonzero, we can choose some y such that   and solve for   in the above expression. Therefore:

 

The final expression must go to zero as   since   and   is continuous. It follows that   is continuous.

Now, we prove that  , for some k, for all positive rational numbers q. Let q=n/m for positive integers n and m. Then

 

by elementary induction on n. Therefore,   and thus

 

for  . Note that if we are restricting ourselves to real-valued  , then   is everywhere positive and so k is real.

Finally, by continuity, since   for all rational x, it must be true for all real x since the closure of the rationals is the reals (that is, we can write any real x as the limit of a sequence of rationals). If   then k = 1. This is equivalent to characterization 1 (or 2, or 3), depending on which equivalent definition of e one uses.

Characterization 2 implies characterization 6Edit

In the sense of definition 2,[1]

 
 
 
 

Characterization 6 implies characterization 4Edit

In the sense of definition 6,   By the way  , therefore definition 6 implies definition 4.

ReferencesEdit

  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition (McGraw–Hill, 1976), chapter 8.
  • Edwin Hewitt and Karl Stromberg, Real and Abstract Analysis (Springer, 1965).