Since all the derivatives of the exponential function are equal to the function itself, we have \[e^{x} = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n-1}}{(n – 1)!} + \frac{x^{n}}{n!} e^{\theta x}\] where \(0 < \theta < 1\). But \(x^{n}/n! \to 0\) as \(n \to \infty\), whatever be the value of \(x\) (Ex. XXVII. 12); and \(e^{\theta x} < e^{x}\). Hence, making \(n\) tend to \(\infty\), we have \[\begin{equation*} e^{x} = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!} + \dots. \tag{1} \end{equation*}\]

The series on the right-hand side of this equation is known as the **exponential series.** In particular we have \[\begin{equation*} e = 1 + 1 + \frac{1}{2!} + \dots + \frac{1}{n!} + \dots; \tag{2} \end{equation*}\] and so \[\begin{equation*} \left(1 + 1 + \frac{1}{2!} + \dots + \frac{1}{n!} + \dots\right)^{x} = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!} + \dots, \tag{3} \end{equation*}\] a result known as the **exponential theorem**. Also \[\begin{equation*} a^{x} = e^{x\log a} = 1 + (x\log a) + \frac{(x\log a)^{2}}{2!} + \dots \tag{4} \end{equation*}\] for all positive values of \(a\).

The reader will observe that the exponential series has the property of reproducing itself when every term is differentiated, and that no other series of powers of \(x\) would possess this property: for some further remarks in this connection see Appendix II.

The power series for \(e^{x}\) is so important that it is worth while to investigate it by an alternative method which does not depend upon Taylor’s Theorem. Let \[E_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!},\] and suppose that \(x > 0\). Then \[\left(1 + \frac{x}{n}\right)^{n} = 1 + n\left(\frac{x}{n}\right) + \frac{n(n – 1)}{1\cdot2} \left(\frac{x}{n}\right)^{2} + \dots + \frac{n(n – 1)\dots 1}{1\cdot2\dots n} \left(\frac{x}{n}\right)^{n}{,}\] which is less than \(E_{n}(x)\). And, provided \(n > x\), we have also, by the binomial theorem for a negative integral exponent, \[\left(1 – \frac{x}{n}\right)^{-n} = 1 + n\left(\frac{x}{n}\right) + \frac{n(n + 1)}{1\cdot2} \left(\frac{x}{n}\right)^{2} + \dots > E_{n}(x).\] Thus \[\left(1 + \frac{x}{n}\right)^{n} < E_{n}(x) < \left(1 – \frac{x}{n}\right)^{-n}.\] But (§ 208) the first and last functions tend to the limit \(e^{x}\) as \(n \to \infty\), and therefore \(E_{n}(x)\) must do the same. From this the equation follows when \(x\) is positive; its truth when \(x\) is negative follows from the fact that the exponential series, as was shown in Ex. LXXXI. 7, satisfies the functional equation \(f(x)f(y) = f(x + y)\), so that \(f(x)f(-x) = f(0) = 1\).

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