|
|
|
|
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.
Characterizations
The five most common definitions of the exponential function exp(x) = ex for real x are:
- 1.
Discussion
Ask a question about 'Characterizations of the exponential function' Start a new discussion about 'Characterizations of the exponential function' Answer questions from other users |
Encyclopedia
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.
Characterizations
The five 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 sum of the infinite series
-
- 3. Define ex to be the unique number y > 0 such that
- 4. Define ex to be the unique solution to the initial value problem
- 5. The exponential function f(x) = ex is the unique Lebesgue-measurable function with f(1) = e that satisfies
- . Alternatively, it is the unique anywhere-continuous function with these properties (Rudin, 1976, chapter 8, exercise 6). As another alternative, it is the only monotonic function satisfying those identities: the identities force f(x) = ex for rational x, and then for irrational x, monotonicity sandwiches f(x) between the values of f(q) for the rational numbers q on each side of x. (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.)
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. 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
Why each characterization makes sense
Each characterization requires some justification to show that it makes sense. For instance, when the value of the function is defined by a sequence or series, the convergence of this sequence or series needs to be established.
Characterization 2
Since
-
it follows from the ratio test that converges for all x.
Characterization 3
Since the integrand is an integrable function of t, the integral expression makes sense. That every real number x corresponds to a unique y > 0 such that
-
follows easily if one can show that 1/t is positive for positive t and that
-
The former is obvious, and the latter follows from the integral test and the divergence of the harmonic series.
Equivalence of the characterizations
The following proof demonstrates the equivalence of the 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. 63-65.
Let be a fixed non-negative real number. Define
By the binomial theorem,
(using 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
Equivalence of characterizations 1 and 3
Here, we define the natural logarithm function in terms of a definite integral as above. By the fundamental theorem of calculus,
-
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 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 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.
|
| |
|
|
