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 $$f(x)f(-x) = f(0) = 1$$.

Example XC
1. Show that $\cosh x = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots,\quad \sinh x = x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots.$

2. If $$x$$ is positive then the greatest term in the exponential series is the $$([x] + 1)$$-th, unless $$x$$ is an integer, when the preceding term is equal to it.

3. Show that $$n! > (n/e)^{n}$$. [For $$n^{n}/n!$$ is one term in the series for $$e^{n}$$.]

4. Prove that $$e^{n} = (n^{n}/n!)(2 + S_{1} + S_{2})$$, where $S_{1} = \frac{1}{1 + \nu} + \frac{1}{(1 + \nu)(1 + 2\nu)} + \dots,\quad S_{2} = (1 – \nu) + (1 – \nu)(1 – 2\nu) + \dots,$ and $$\nu = 1/n$$; and deduce that $$n!$$ lies between $$2(n/e)^{n}$$ and $$2(n + 1)(n/e)^{n}$$.

5. Employ the exponential series to prove that $$e^{x}$$ tends to infinity more rapidly than any power of $$x$$. [Use the inequality $$e^{x} > x^{n}/n!$$.]

6. Show that $$e$$ is not a rational number. [If $$e = p/q$$, where $$p$$ and $$q$$ are integers, we must have $\frac{p}{q} = {1 + {}} 1 + \frac{1}{2!}+\frac{1}{3!} + \dots + \frac{1}{q!} + \dots$ or, multiplying up by $$q!$$, $q! \left(\frac{p}{q} – 1 – 1 – \frac{1}{2!} – \dots – \frac{1}{q!}\right) = \frac{1}{q + 1} + \frac{1}{(q + 1)(q + 2)} + \dots$ and this is absurd, since the left-hand side is integral, and the right-hand side less than $$\{1/(q + 1)\} + \{1/(q + 1)\}^{2} + \dots = 1/q$$.]

7. Sum the series $$\sum\limits_{0}^{\infty} P_{r}(n)\dfrac{x^{n}}{n!}$$, where $$P_{r}(n)$$ is a polynomial of degree $$r$$ in $$n$$. [We can express $$P_{r}(n)$$ in the form $A_{0} + A_{1}n + A_{2}n(n – 1) + \dots + A_{r}n(n – 1) \dots (n – r + 1),$ and \begin{aligned} \sum_{0}^{\infty} P_{r}(n) \frac{x^{n}}{n!} &= A_{0}\sum_{0}^{\infty}\frac{x^{n}}{n!} + A_{1}\sum_{1}^{\infty}\frac{x^{n}}{(n – 1)!} + \dots + A_{r}\sum_{r}^{\infty}\frac{x^{n}}{(n – r)!}\\ &= (A_{0} + A_{1}x + A_{2}x^{2} + \dots + A_{r}x^{r})e^{x}.]\end{aligned}

8. Show that $\sum_{1}^{\infty} \frac{n^{3}}{n!} x^{n} = (x + 3x^{2} + x^{3})e^{x},\quad \sum_{1}^{\infty} \frac{n^{4}}{n!} x^{n} = (x + 7x^{2} + 6x^{3} + x^{4})e^{x};$ and that if $$S_{n} = 1^{3} + 2^{3} + \dots + n^{3}$$ then $\sum_{1}^{\infty} S_{n}\frac{x^{n}}{n!} = \tfrac{1}{4}(4x + 14x^{2} + 8x^{3} + x^{4})e^{x}.$ In particular the last series is equal to zero when $$x = -2$$.

9. Prove that $$\sum (n/n!) = e$$, $$\sum (n^{2}/n!) = 2e$$, $$\sum (n^{3}/n!) = 5e$$, and that $$\sum (n^{k}/n!)$$, where $$k$$ is any positive integer, is a positive integral multiple of $$e$$.

10. Prove that $$\sum\limits_{1}^{\infty} \dfrac{(n – 1)x^{n}}{(n + 2)n!} = \left\{(x^{2} – 3x + 3)e^{x} + \frac{1}{2}x^{2} – 3\right\}/x^{2}$$.

[Multiply numerator and denominator by $$n + 1$$, and proceed as in Ex. 7.]

11. Determine $$a$$, $$b$$, $$c$$ so that $$\{(x + a)e^{x} + (bx + c)\}/x^{3}$$ tends to a limit as $$x \to 0$$, evaluate the limit, and draw the graph of the function $$e^{x} + \dfrac{bx + c}{x + a}$$.

12. Draw the graphs of $$1 + x$$, $$1 + x + \frac{1}{2}x^{2}$$, $$1 + x + \frac{1}{2}x^{2} + \frac{1}{6}x^{3}$$, and compare them with that of $$e^{x}$$.

13. Prove that $$e^{-x} – 1 + x – \dfrac{x^{n}}{2!} + \dots – (-1)^{n}\dfrac{x^{n}}{n!}$$ is positive or negative according as $$n$$ is odd or even. Deduce the exponential theorem.

14. If $X_{0} = e^{x},\quad X_{1} = e^{x} – 1,\quad X_{2} = e^{x} – 1 – x,\quad X_{3} = e^{x} – 1 – x – (x^{2}/2!),\ \dots,$ then $$dX_{\nu}/dx = X_{\nu-1}$$. Hence prove that if $$t > 0$$ then $X_{1}(t) = \int_{0}^{t} X_{0}\, dx < te^{t},\quad X_{2}(t) = \int_{0}^{t} X_{1}\, dx < \int_{0}^{t} xe^{x}\, dx < e^{t} \int_{0}^{t} x\, dx = \frac{t^{2}}{2!} e^{t},$ and generally $$X_{\nu}(t) < \dfrac{t^{\nu}}{\nu!} e^{t}$$. Deduce the exponential theorem.

15. Show that the expansion in powers of $$p$$ of the positive root of $$x^{2+p} = a^{2}$$ begins with the terms $a\{1 – \tfrac{1}{2} p\log a + \tfrac{1}{8} p^{2}\log a (2 + \log a)\}.$