Characterizations of the exponential function

Characterizations

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

1. Define e^x by the limit 2. Define e^x as the value of the infinite series 3. Define e^x to be the unique number y > 0 such that 4. Define e^x to be the unique solution to the initial value problem 5. The exponential function f(x) = e^x 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 x at which f ( x ) is continuous. As shown below, if f ( x + y ) = f ( x ) f ( y ) for all x and y and f ( x ) is continuous at any single point x then f ( x ) 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) = e^x is guaranteed for rational x by the above properties (see below), one could also use monotonicity or other properties to enforce the choice of e^x for irrational x, but such alternatives appear to be uncommon. 6. Let e be the unique real number satisfying

This number can be shown to exist and be unique. This definition is particularly suited to computing the derivative of the exponential function. Then define e^x 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

f ( x + i y ) = e x ( cos ⁡ ( 2 y ) + i sin ⁡ ( 2 y ) ) = e x + 2 i y

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

f ( i ) = cos ⁡ ( 1 ) + i sin ⁡ ( 1 ) .

In particular, the alternate condition in (5) that f ′ ( 0 ) = 1 is sufficient since it implicitly stipulates that f be conformal.

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

lim n → ∞ | x n + 1 / ( n + 1 ) ! x n / n ! | = lim n → ∞ | x n + 1 | = 0 < 1.

it follows from the ratio test that ∑ n = 0 ∞ x n n ! converges for all x.

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 R + to R defined by

∫ 1 ( ⋅ ) d t t

is a bijection. As t − 1 is positive for positive t, this function is monotone increasing, hence one-to-one. If the two integrals

∫ 1 ∞ d t t = ∞ ∫ 1 0 d t t = − ∞

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 x ≥ 0 be a fixed non-negative real number. Define

s n = ∑ k = 0 n x k k ! ,   t n = ( 1 + x n ) n .

By the binomial theorem,

t n = ∑ k = 0 n ( n k ) x k n k = 1 + x + ∑ k = 2 n n ( n − 1 ) ( n − 2 ) ⋯ ( n − ( k − 1 ) ) x k k ! n k = 1 + x + x 2 2 ! ( 1 − 1 n ) + x 3 3 ! ( 1 − 1 n ) ( 1 − 2 n ) + ⋯ ⋯ + x n n ! ( 1 − 1 n ) ⋯ ( 1 − n − 1 n ) ≤ s n

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

lim sup n → ∞ t n ≤ lim sup n → ∞ s n = e x

where e^x is in the sense of definition 2. Here, we must use limsups, because we don't yet know that t_n actually converges. Now, for the other direction, note that by the above expression of t_n, if 2 ≤ m ≤ n, we have

1 + x + x 2 2 ! ( 1 − 1 n ) + ⋯ + x m m ! ( 1 − 1 n ) ( 1 − 2 n ) ⋯ ( 1 − m − 1 n ) ≤ t n .

Fix m, and let n approach infinity. We get

s m = 1 + x + x 2 2 ! + ⋯ + x m m ! ≤ lim inf n → ∞ t n

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

lim sup n → ∞ t n ≤ e x ≤ lim inf n → ∞ t n

so that

lim n → ∞ t n = e x .

We can then extend this equivalence to the negative real numbers by noting ( 1 − r n ) n ( 1 + r n ) n = ( 1 − r 2 n 2 ) n and taking the limit as n goes to infinity.

The error term of this limit-expression is described by

( 1 + x n ) n = e x ( 1 − x 2 2 n + x 3 ( 8 + 3 x ) 24 n 2 + ⋯ ) ,

where the polynomial's degree (in x) in the term with denominator n^k 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,

d d x ln ⁡ x = d d x ∫ 1 x 1 t d t = 1 x .

Besides, ln ⁡ 1 = ∫ 1 1 1 t d t = 0

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

y = lim n → ∞ ( 1 + x n ) n .

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

ln ⁡ y = ln ⁡ lim n → ∞ ( 1 + x n ) n = lim n → ∞ ln ⁡ ( 1 + x n ) n .

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

ln ⁡ y = lim n → ∞ n ln ⁡ ( 1 + x n ) = lim n → ∞ x ln ⁡ ( 1 + ( x / n ) ) ( x / n ) .

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 a^b can be defined for real b as e^{b lna}.)

= x ⋅ lim h → 0 ln ⁡ ( 1 + h ) h  where  h = x n = x ⋅ lim h → 0 ln ⁡ ( 1 + h ) − ln ⁡ 1 h = x ⋅ d d t ln ⁡ t | t = 1 = x .

Equivalence of characterizations 2 and 4

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

Therefore d n y d x n | x = 0 = y ( 0 ) = 1.

Using Taylor series, y = ∑ n = 0 ∞ f ( n ) ( 0 ) n ! x n = ∑ n = 0 ∞ 1 n ! x n = ∑ n = 0 ∞ x n n ! . This shows that definition 4 implies definition 2.

In the sense of definition 2,

d d x e x = d d x ( 1 + ∑ n = 1 ∞ x n n ! ) = ∑ n = 1 ∞ n x n − 1 n ! = ∑ n = 1 ∞ x n − 1 ( n − 1 ) ! = ∑ k = 0 ∞ x k k ! ,  where  k = n − 1 = e x

Besides, e 0 = 1 + 0 + 0 2 2 ! + 0 3 3 ! + ⋯ = 1. This shows that definition 2 implies definition 4.

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 f ( x ) satisfying f ( x + y ) = f ( x ) f ( y ) , and then one proves that continuity implies f ( x ) = e k x for some k, and finally f ( 1 ) = e implies k=1.

First, we prove a few elementary properties from f ( x ) satisfying f ( x + y ) = f ( x ) f ( y ) and the assumption that f ( x ) is not identically zero:

If f ( x ) is nonzero anywhere (say at x=y), then it is non-zero everywhere. Proof: f ( y ) = f ( x ) f ( y − x ) ≠ 0 implies f ( x ) ≠ 0 .

f ( 0 ) = 1 . Proof: f ( x ) = f ( x + 0 ) = f ( x ) f ( 0 ) and f ( x ) is non-zero.

f ( − x ) = 1 / f ( x ) . Proof: 1 = f ( 0 ) = f ( x − x ) = f ( x ) f ( − x ) .

If f ( x ) is continuous anywhere (say at x = y), then it is continuous everywhere. Proof: f ( x + δ ) − f ( x ) = f ( x − y ) [ f ( y + δ ) − f ( y ) ] → 0 as δ → 0 by continuity at y.

The second and third properties mean that it is sufficient to prove f ( x ) = e x for positive x.

If f ( x ) is a Lebesgue-integrable function, then we can define

g ( x ) = ∫ 0 x f ( x ′ ) d x ′ .

It then follows that

g ( x + y ) − g ( x ) = ∫ x x + y f ( x ′ ) d x ′ = ∫ 0 y f ( x + x ′ ) d x ′ = f ( x ) g ( y ) .

Since f ( x ) is nonzero, we can choose some y such that g ( y ) ≠ 0 and solve for f ( x ) in the above expression. Therefore:

f ( x + δ ) − f ( x ) = [ g ( x + δ + y ) − g ( x + δ ) ] − [ g ( x + y ) − g ( x ) ] g ( y ) = [ g ( x + y + δ ) − g ( x + y ) ] − [ g ( x + δ ) − g ( x ) ] g ( y ) = f ( x + y ) g ( δ ) − f ( x ) g ( δ ) g ( y ) = g ( δ ) f ( x + y ) − f ( x ) g ( y ) .

The final expression must go to zero as δ → 0 since g ( 0 ) = 0 and g ( x ) is continuous. It follows that f ( x ) is continuous.

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

f ( n m ) = f ( 1 m + ⋯ + 1 m ) = f ( 1 m ) n

by elementary induction on n. Therefore, f ( 1 / m ) m = f ( 1 ) and thus

f ( n m ) = f ( 1 ) n / m = e k ( n / m ) .

for k = ln ⁡ [ f ( 1 ) ] . Note that if we are restricting ourselves to real-valued f ( x ) , then f ( x ) = f ( x / 2 ) 2 is everywhere positive and so k is real.

Finally, by continuity, since f ( x ) = e k x 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 f ( 1 ) = e 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 6

In the sense of definition 2,

= lim h → 0 1 h ( ( 1 + h + h 2 2 ! + h 3 3 ! + h 4 4 ! + ⋯ ) − 1 ) = lim h → 0 ( 1 + h 2 ! + h 2 3 ! + h 3 4 ! + ⋯ ) = 1

Characterization 6 implies characterization 4

In the sense of definition 6, d d x e x = lim h → 0 e x + h − e x h = e x ⋅ lim h → 0 e h − 1 h = e x . By the way e 0 = 1 , therefore definition 6 implies definition 4.