Characterizations of the exponential function

In mathematics, the exponential function can be characterized in many ways. The following three 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

1 Characterizations 2 Why each characterization makes sense 2.1 Characterization 1 2.2 Characterization 2 2.3 Characterization 3 3 Equivalence of the characterizations 3.1 Equivalence of characterizations 1 and 2 3.2 Equivalence of characterizations 1 and 3 4 References

Characterizations

The four most common definitions of the exponential function exp(x) = e^x are the following.

1. Define e^x by the limit

<math>e^x = \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n.<math>

2. Define e^x as the sum of the infinite series

<math>e^x = \sum_{n=0}^\infty {x^n \over n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots<math>

Here, n! stands for the factorial of n.

3. Define e^x to be the unique number y > 0 such that

<math>\int_{1}^{y} \frac{dt}{t} = x.<math>

4. Define e^x to be the unique solution to the initial value problem

<math>y'=y,\quad y(0)=1.<math>

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 1

It can be shown that the sequence

<math>a_n = \left(1+\frac{x}{n}\right)^n<math>

is an increasing sequence which is bounded above. Since every bounded, increasing sequence of real numbers converges to a unique real number, this characterization makes sense.

Characterization 2

To show the infinite series converges at x = 1, it is enough to compare with a geometric series:

<math>1 + 1 + {1 \over 2!} + {1 \over 3!} + {1 \over 4!} + \cdots<math>

<math>\le 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots = 3.<math>

To show that the series converges for all x, we use the ratio test, which shows that the series has an infinite radius of convergence, since

<math>\lim_{n\to\infty} \left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|

  = \lim_{n\to\infty} \left|\frac{x}{n+1}\right|
  = 0 < 1 \mbox{ for all }x \mbox{.}<math>

Characterization 3

In this case, we define the natural logarithm function ln(x) first, and then define exp(x) as the inverse of the natural logarithm. In other words, for all y > 0, define

<math>\ln (y) = \int_{1}^{y} \frac{1}{t} \, dt.<math>

Since 1/t is continuous for all t > 0, this function makes sense, and since 1/t is positive for all t > 0, this function is strictly increasing (hence, injective) for y > 0. (Note that if y < 1, then ln(y) is a negative number.) By the integral test and the divergence of the harmonic series, it follows that ln(y) → ∞ as y → ∞. By a similar argument, a change of variables (t <math>\mapsto<math> 1/t) shows that ln(y) → −∞ as y → 0. To sum up, ln(y) maps the infinite interval (0, ∞) bijectively onto the real line (−∞, ∞). Hence, for any real number x, there must exist a unique number y > 1 such that ln(y) = x.

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. The irrationality of e follows quickly from this proof, see theorem 3.32.

Let x be a fixed real number. Define

<math>s_n = \sum_{k=0}^n\frac{x^k}{k!},\ t_n=\left(1+\frac{x}{n}\right)^n.<math>

By the binomial theorem,

<math>t_n=\sum_{k=0}^n{n \choose k}\frac{x^k}{n^k}=1+x+\sum_{k=2}^n\frac{n(n-1)(n-2)\cdots(n-(k-1))x^k}{k!\,n^k}<math>

<math>=1+x+\frac{x^2}{2!}\left(1-\frac{1}{n}\right)+\frac{x^3}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\cdots+\frac{x^n}{n!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)\le s_n<math>

so that

<math>\limsup_{n\to\infty}t_n \le \limsup_{n\to\infty}s_n = e^x<math>

where e^x is in the sense of definition 2. Here, we must use limsup's, 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

<math>1+x+\frac{x^2}{2!}\left(1-\frac{1}{n}\right)+\cdots+\frac{x^m}{m!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\le t_n.<math>

Fix m, and let n approach infinity. We get

<math>s_m = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^m}{m!} \le \liminf_{n\to\infty}t_n<math>

(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

<math>\limsup_{n\to\infty}t_n \le e^x \le \liminf_{n\to\infty}t_n \le \limsup_{n\to\infty}t_n<math>

so that

<math>\lim_{n\to\infty}t_n = e^x.<math>

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,

<math>\frac{d}{dx}\left( \ln x \right) = \frac{1}{x}.<math>

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

<math>y=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n.<math>

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

<math>\ln y=\ln\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}\ln\left(1+\frac{x}{n}\right)^n.<math>

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

<math>\ln y=\lim_{n\to\infty}n\ln \left(1+\frac{x}{n}\right)=\lim_{n\to\infty}\frac{x\ln\left(1+(x/n)\right)}{(x/n)}<math>

<math>=x\cdot\lim_{h\to 0}\frac{\ln\left(1+h\right)}{h} \quad \mbox{ where }h=\frac{x}{n}<math>

<math>=x\cdot\frac{d}{dt}\left( \ln t\right) \quad \mbox{ at }t=1<math>

<math>=x\cdot\frac{1}{t} \quad \mbox{ at }t=1<math>

<math>\!\, = x.<math>

References

• Principles of Mathematical Analysis, Walter Rudin, 3rd edition, McGraw-Hill, 1976.