§ 2. Expansions of the exponential and circular functions, and of the functions associated with them

1. Expansion of the real exponential

Since $\mathrm{D}^n \left( e^x \right) = e^x$ the Taylor expansion of order $n$ for $e^x$ is

The remainder in this formula is $> 0$ for $x > 0$ and has the sign of $(-1)^{n+1}$ when $x < 0$; moreover, the inequality of the mean shows that

Now one knows that the sequence $\left( x^n / n! \right)$ has limit 0 when $n$ increases indefinitely, for all $x \geqslant 0$ (Gen. Top., IV, p. 365); thus, keeping $x$ fixed and letting $n$ grow indefinitely in (1) it follows, from (2) and (3), that

and the series on the right-hand side is absolutely and uniformly convergent on every compact interval in $\mathbf{R}$. In particular, one has the formula

This formula allows us to calculate rational approximations as close as we desire to the number $e$; one obtains

to within $1/10^9$. Formula (5) proves, moreover, that $e$ is an irrational number $^{2}$ (Gen. Top., IV, p. 375).

2. Expansions of the complex exponential, of $\cos x$ and $\sin x$

Let $z$ be an arbitrary complex number and consider the function $\varphi(t) = e^{zt}$ of the real variable $t$; we have $D^n \varphi(t) = z^n e^{zt}$ and $e^z = \varphi(1)$; expressing $\varphi(1)$ by means of its Taylor series of order $n$ about the point $t = 0$ (II, p. 62) thus gives

a formula which is equivalent to (1) when $z$ is real. The remainder

in this formula can be bounded above, in absolute value, by using the inequality of the mean; if $z = x + i y$ we have $|e^{zt}| = e^{xt}$, so $|e^{zt}| \leq 1$ if $x \leq 0$, $|e^{zt}| \leq e^x$ if $x > 0$; hence

As above we conclude that

the series being absolutely and uniformly convergent on every compact subset of $\mathbf{C}$.

From (6) one deduces in particular that

from which we deduce the Taylor expansions of $\cos x$ and $\sin x$; on taking the real part of (10) for order $2n+1$ we have

with remainder bounded by

Similarly, taking the imaginary part of (10) for order $2n$, we obtain

with remainder bounded by

Moreover, on comparing the remainders in (11) for orders $2n+1$ and $2n+3$, we have

and taking (12) into account we see that the remainder in (11) has the same sign as $(-1)^{n+1}$ no matter what $x$; in the same way we can show that the remainder in (13) has the same sign as $(-1)^n x$. In particular, for $n=0$ and $n=1$ in (11), and for $n=1$ and $n=2$ in (13) we obtain the inequalities

Finally, on putting $z = i x$ in (9) we have

these series being absolutely and uniformly convergent on every compact interval.

It is clear, furthermore, that the formulae (17) and (18) remain valid for every complex $x$, the series on the right-hand side being absolutely and uniformly convergent on every compact subset of $\mathbf{C}$. In particular, for every $x$ (real or complex)

3. The binomial expansion

Let $m$ be an arbitrary real number. For $x > 0$ we have

the Taylor formula of order $n$ about the point $x = 0$ for the function $(1 + x)^m$ shows that for every $x > -1$

with

where we put $\binom{m}{n} = \frac{m(m-1)\ldots(m-n+1)}{n!}$. The formula (19) reduces to the binomial formula (Alg., I, p. 99) when $m$ is an integer $> 0$ and $n \geq m$; by extension, we again call it the binomial formula, and the coefficients $\binom{m}{n}$ are called the binomial coefficients, when $m$ is an arbitrary real number and $n$ is an arbitrary integer $> 0$.

The remainder in (19) has the same sign as $\binom{m}{n+1}$ if $x > 0$, and the sign of $(-1)^{n+1} \binom{m}{n+1}$ if $-1 < x < 0$. Since $\left| \frac{x-t}{1+t} \right| \leq |x|$ for $t > -1$ in the interval with endpoints 0 and $x$, we have the following bound for the remainder, for $m$ and $n$ arbitrary and $x > -1$:

If we suppose $x \geq 0$, and $n \geq m-1$, then $(1+t)^{n-m+1} \geq 1$ on the interval of integration, so

which gives the estimate

for the remainder. On the other hand, suppose that $-1 \leq m < 0$; if one makes the change of variable $u = \frac{x-t}{x(1+t)}$ in the integral (19) one obtains

To estimate the integral for $x > -1$ we remark that, since $m + 1 < 1$, the integral $\int_0^1 \frac{u^n \, du}{(1-u)^{m+1}}$ converges and bounds the right-hand side of (22) since $1 + u x > 1 - u$. Now, for $-1 < x < 0$ the hypothesis on $m$ implies that all the terms $\binom{m}{1} x, \binom{m}{2} x^2, \ldots, \binom{m}{n} x^n$ which appear in the right-hand side of (19) are $\geq 0$, and hence $r_n(x) \leq (1 + x)^m$, from which, on dividing by $(1 + x)^m$,

Moreover, for $-1 < x < 0$ the factor in front of the integral is $\geq 0$, so, letting $x$ approach $-1$,

and consequently for $-1 \leq m < 0$ and $x > -1$ we have

From these inequalities we can, for a start, deduce that for $|x| < 1$ we have

the right-hand side (called the binomial series) being absolutely and uniformly convergent on every compact subset of $]-1, +1[$. Indeed one can write

whence

If $|x| \leq r < 1$ there is an $n_0$ such that $1 + \frac{|m|}{n_0} < \frac{1}{r'}$, where $r < r' < 1$; whence, putting

we have

which proves the proposition. On the other hand, for $x > 1$, the absolute value of the general term of the series (24) increases indefinitely with $n$ if $m$ is not an integer $\geq 0$; indeed, from (25), we have for $n > n_1 \geq |m+1|$

Let $n_0 \geq n_1$ be such that for $n \geq n_0$ we have $1 - \frac{|m+1|}{n} > \frac{1}{x'}$, where $1 < x' < x$. If we put

We remark that for $m = -1$ the algebraic identity

In the second place let us study the convergence of the binomial series for $x = 1$ or $x = -1$ (excluding the trivial case $m = 0$):

$a) \ m \leq -1$. The product with general term $1 - \frac{m+1}{n}$ converges to $+\infty$ if $m < -1$, to 1 if $m = -1$, so it follows from (25) that for $x = \pm 1$ the general term of the binomial series does not tend to 0.

$b) \ -1 < m < 0$. This time the product with general term $1 - \frac{m+1}{n}$ converges to 0, so the inequality (21) shows that $r_n(1)$ tends to 0. Thus the binomial series converges for $x = 1$ and has sum $2^m$; moreover, the binomial series is uniformly convergent on every interval $]x_0, 1]$ with $-1 < x_0 \leq 1$, by virtue of what we saw above and of (21). On the other hand, for $x = -1$ all the terms on the right-hand side of (24) are $\geq 0$, if this series were convergent one could deduce that the binomial series would be normally convergent on $[-1, 1]$ and so would have for its sum a continuous function on this interval, which is absurd because $(1+x)^m$ is not bounded on $[-1, 1]$ for $m < 0$. We conclude that also for $x = 1$ the binomial series is not absolutely convergent.

$c) \ m > 0$. The definition of $r_n(x)$ shows that $r_n(x)$ tends to the limit $r_n(-1)$ when $x$ tends to $-1$; on passing to the limit in (20) one concludes that $|r_n(-1)| \leq \left| \binom{m-1}{n} \right|$, and since $m-1 > -1$ we see that for $x = -1$ the binomial series is convergent. Furthermore, for $n > m+1$ all the terms of this series have the same sign; thus the binomial series is normally convergent on the interval [−1, 1] and has sum (1 + x)^m on this interval.

4. EXPANSIONS OF log(1 + x), OF Arc tan x AND OF Arc sin x

Let us integrate the two sides of (26) between 0 and x; we obtain the Taylor expansion of order n of log(1 + x), valid for x > −1

The remainder has the same sign as (−1)^n if x > 0, and is < 0 if −1 < x < 0; further, when x > 0, we have 1 + t ≥ 1 for 0 ≤ t ≤ x, and, when −1 < x < 0, we have 1 + t ≥ 1 − |x| for x ≤ 0; whence the estimates for the remainder

From these last two formulae one deduces immediately that for −1 < x ≤ 1 one has

the series being uniformly convergent on every compact interval contained in ]−1, +1], and absolutely convergent for |x| < 1.

On the other hand, for |x| > 1 the general term in the series on the right-hand side of (30) increases indefinitely in size with n (III, p. 106). For x = −1 the series reduces to the harmonic series, which has sum +∞ (Gen. Top., IV, p. 365).

Similarly, let us replace x by x^2 in (26) and integrate both sides between 0 and x; we obtain the Taylor expansion of order 2n − 1 for Arc tan x, valid for all real x

The remainder has the sign of (−1)^n x, and since 1 + t^2 ≥ 1 for all t we have the estimate

from which one deduces that, for |x| ≤ 1,

the series being uniformly convergent on [−1, +1], and absolutely convergent for |x| < 1.

In particular, for x = 1 one obtains the formula

For |x| > 1 the general term of the right-hand side of (33) increases indefinitely in size with n.

Finally, for the Taylor expansion of Arc sin x we start from the expansion of its derivative (1 − x^2)^{-1/2}; this last expansion is obtained by replacing x by −x^2 in the expansion of (1 + x)^{-1/2} as a binomial series; for |x| < 1 this gives

with, by (23), the bound

for the remainder.

On taking the primitive of the preceding expansion we obtain

where R_n(x) has the sign of x and satisfies the inequality

Further, the relation (35) shows that R_n(x) tends to a limit when x approaches 1 or −1, so one has

But the right-hand side of (37) tends to 0 when n tends to +∞ : for, since the integral $\int_0^1 dt / \sqrt{1-t^2}$ is convergent, for every ε > 0 there is an a such that 0 < a < 1 and $\int_a^1 dt / \sqrt{1-t^2} \leq \varepsilon$; on the other hand we have

and so there is an n_0 such that for n ≥ n_0 one has $\frac{a^{2n+3}}{(2n+3)\sqrt{1-a^2}} \leq \varepsilon$, whence, finally, |R_n(x)| ≤ 2ε for |x| ≤ 1 and n ≥ n_0. Thus one has

(38)

the right-hand side being normally convergent on the compact interval [−1, 1].

In the opposite case one can show, as for the binomial series, that the general term in the series on the right-hand side of (38) increases indefinitely in absolute value for |x| > 1. On putting $x = \frac{1}{2}$, for example, in (38) we obtain a new expression for the number $\pi$:

which is much better suited than formula (34) to calculating approximations to $\pi$ (see Calcul numérique); one thus obtains

accurate to within $1/10^9$.³

³ The number $\pi$ is not only irrational (cf. III, p. 126, exerc. 5) but even transcendental over the field $\mathbf{Q}$ of rational numbers, as was shown for the first time in 1882 by Lindemann (see for example D Hilbert, Gesammelte Abhandlungen, v. 1, p. 1, Berlin (Springer), 1932).

Exercises

See the exercises for § 2.