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 \[a\{1 – \tfrac{1}{2} p\log a + \tfrac{1}{8} p^{2}\log a (2 + \log a)\}.\]


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