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
- Characterizations
- Larger domains
- Proof that each characterization makes sense
- Characterization 2
- Characterization 3
- Equivalence of the characterizations
- Equivalence of characterizations 1 and 2
- Equivalence of characterizations 1 and 3
- Equivalence of characterizations 2 and 4
- Equivalence of characterizations 1 and 5
- Characterization 2 implies characterization 6
- Characterization 6 implies characterization 4
- References
Characterizations
The six most common definitions of the exponential function exp(x) = ex for real x are:
1. Define ex by the limit2. Define ex as the value of the infinite series3. Define ex to be the unique number y > 0 such that4. Define ex to be the unique solution to the initial value problem5. 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 pointThis number can be shown to exist and be unique. 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 domains
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
Proof that each characterization makes sense
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 2
Since
it follows from the ratio test that
Characterization 3
Since the integrand is an integrable function of t, the integral expression is well-defined. Now we must show that the function from
is a bijection. As
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 characterizations
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 2
The following argument is adapted from a proof in Rudin, theorem 3.31, p. 63–-5.
Let
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 ≤ m ≤ n, 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
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 3
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 4
Let n be a non-negative integer. In the sense of definition 4 and by induction,
Therefore
Using Taylor series,
In the sense of definition 2,
Besides,
Equivalence of characterizations 1 and 5
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
First, we prove a few elementary properties from
The second and third properties mean that it is sufficient to prove
If
It then follows that
Since
The final expression must go to zero as
Now, we prove that
by elementary induction on n. Therefore,
for
Finally, by continuity, since
Characterization 2 implies characterization 6
In the sense of definition 2,
Characterization 6 implies characterization 4
In the sense of definition 6,