§2. Series expansions
347
One might do without all this by starting from the functional equation
for the series exp and reconstituting it all by this route. In fact, we could
have done this from the end of Chap. II, but it was already quite long ...
Let us now pretend not to have read nOs 1 to 9 of this Chap. IV and let
us start from the relation
(10.1)
exp(x). exp(y) = exp(x + y),
valid for all x, y E C. It has immediate consequences.
First, (i) the function exp(z) never vanishes, even for z complex, (ii) exp(x)
is strictly positive for x real, (iii) the function exp is strictly increasing, since
the series expx = 1 + x/I! + x 2 /2! + ... shows that x > 0 ==} expx > 1
(iv) we have
exp(px/q)q = exp(x)P
for x E C and p,q integers. In particular, exp(px/q) = exp(x)p/q for x real.
It is unnecessary to detail these obvious properties again.
Secondly, the exponential function is analytic on C, not only by reason of
the general result of Chap. II, nO 19, but, more simply, because for any a E C
it has a power series expansion
expz = exp(a) exp(z - a) = exp(a) ~)z - a)n/nL
In particular, the function exp is continuous on C, so on JRj since it takes
arbitrarily large values on JR, [exp n = exp(l)n with exp 1 > 1] and arbitrarily
close to 0 [since exp( -x) = 1/ expx], and since it is strictly increasing, the
function exp maps JR bijectively onto JR+.
Thirdly, the exponential function is identical to its derivative, and so to
all its successive derivatives as shown by Chap. II, nO 19, on the derived series
defined by the general algorithm or as the traditional limit of the quotient.
This result applies even for complex z.
An obvious calculation shows more generally that, for t E C, the function
f(z) = exp(tz) satisfies f'(z) = tf(z)
Le. is proportional to its derivative, whence
for any n E No This allows us to solve "linear differential equations with
constant coefficients". Suppose for example that we want to find the (or, at
this level of the exposition, some) functions y = f(x) satisfying
y"" - 3y'" + 3y" - 3y' + 2y = o.
We are tempted to look for solutions of the form exp(tx)j the parameter t
must clearly verify the algebraic equation
347
One might do without all this by starting from the functional equation
for the series exp and reconstituting it all by this route. In fact, we could
have done this from the end of Chap. II, but it was already quite long ...
Let us now pretend not to have read nOs 1 to 9 of this Chap. IV and let
us start from the relation
(10.1)
exp(x). exp(y) = exp(x + y),
valid for all x, y E C. It has immediate consequences.
First, (i) the function exp(z) never vanishes, even for z complex, (ii) exp(x)
is strictly positive for x real, (iii) the function exp is strictly increasing, since
the series expx = 1 + x/I! + x 2 /2! + ... shows that x > 0 ==} expx > 1
(iv) we have
exp(px/q)q = exp(x)P
for x E C and p,q integers. In particular, exp(px/q) = exp(x)p/q for x real.
It is unnecessary to detail these obvious properties again.
Secondly, the exponential function is analytic on C, not only by reason of
the general result of Chap. II, nO 19, but, more simply, because for any a E C
it has a power series expansion
expz = exp(a) exp(z - a) = exp(a) ~)z - a)n/nL
In particular, the function exp is continuous on C, so on JRj since it takes
arbitrarily large values on JR, [exp n = exp(l)n with exp 1 > 1] and arbitrarily
close to 0 [since exp( -x) = 1/ expx], and since it is strictly increasing, the
function exp maps JR bijectively onto JR+.
Thirdly, the exponential function is identical to its derivative, and so to
all its successive derivatives as shown by Chap. II, nO 19, on the derived series
defined by the general algorithm or as the traditional limit of the quotient.
This result applies even for complex z.
An obvious calculation shows more generally that, for t E C, the function
f(z) = exp(tz) satisfies f'(z) = tf(z)
Le. is proportional to its derivative, whence
for any n E No This allows us to solve "linear differential equations with
constant coefficients". Suppose for example that we want to find the (or, at
this level of the exposition, some) functions y = f(x) satisfying
y"" - 3y'" + 3y" - 3y' + 2y = o.
We are tempted to look for solutions of the form exp(tx)j the parameter t
must clearly verify the algebraic equation