Value of "e" in the C log() function

[JosephKK]

Note the crucial words "complex-valued function defined on an open set S
in the complex plane". For complex functions, being differentiable on an
open disc is equivalent to being given by a convergent power series
there, so authors are free to use whichever definition they find most
convenient.

For (real valued) functions of a *real* variable, the two conditions are
no longer equivalent - a famous counterexample has been provided at
least twice already in this thread - and "analytic" is conventionally
reserved for the stronger condition.

Is this tantamount to saying that the definition of "analytic" has
changed (in current undergraduate texts) in the last 20 years?
.

 

[JosephKK]

Okay, perhaps you should propose a definition for exp(x), and we can see
how the desired properties follow.


exp(-x) is a notational convenience. You can replace it with 1/exp(x)
and the proof will go through the same way, using the quotient rule
instead of the product rule. Does that make you happier?


I don't think that it does.

Here is another theorem, which doesn't mention exp(x) at all.

Lemma. Suppose g is a once-differentiable function on R such that
g'(x) = g(x) for all x in R. Then either g is the zero function, or
g(x) is nonzero for all x in R.

Proof. Suppose, in order to get a contradiction, that there exist x0,
x1 such that g(x0) = 0, g(x1) != 0.

Suppose first that x0 < x1 and g(x1) > 0. Let y = sup { x < x1 : g(x) = 0 }.
Because g is differentiable, g is continuous; it follows that g(y) = 0
and g > 0 on the (nonempty) interval (y, x1). Since g = g', we have
that g is strictly increasing on (y, x1). Choose y1 in (y, x1) such
that 0 < y-y1 < 1. By the mean value theorem, there exists x in (y, y1)
such that g'(x) = g(y1)/(y-y1) > g(y1). Since g(x)=g'(x) we have g(x) >
g(y1). But this is absurd since x < y1 and g is increasing on (y, x1).

Suppose next that x0 > x1 and g(x1) > 0. Let y = inf { x > x1 : g(x) = 0 }.
As before, g(y) = 0 and g > 0 on (x1, y). Thus g is increasing on (x1, y).
But g(x1) > 0 and g(y) = 0 so this is absurd.

The cases where g(x1) < 0 are similar and left as an exercise to the
reader. QED.

Theorem. Suppose f,g are two once-differentiable functions on R such
that f'(x) = f(x) and g'(x) = g(x) for all x in R. Suppose further that
g is not the zero function. Then there exists a constant c such that
f(x) = c g(x) for all x in R.

Proof. By the lemma, g(x) is never zero, so the function h(x)=f(x)/g(x)
is differentiable on R. By the quotient rule, for any x in R we have

h'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)^2
= (f(x)g(x) - f(x)g(x))/g(x)^2
= 0

since f'(x)=f(x), g'(x)=g(x). By the mean value theorem h is constant,
i.e. there exists c such that h(x)=c for all x. Thus f(x) = c g(x) for
all x in R. QED.

With this theorem in hand, start with your favorite definition for
exp(x). Use it to prove that exp'(x) = exp(x) for all x, and that
exp(x) is not the zero function. Then take g(x)=exp(x) in the theorem;
it follows that if f is any other function with f(x)=f'(x), then
f(x) = c*exp(x) for all x.

If you can also prove from your definition that exp(x) is never zero,
you can dispense with the lemma, which is most of the work.

Thank you for that "tour de force". I useful side effect of this is
that on any complex surface S in C, logarithms to any other base are
related ln() by a simple scalar (possibly complex) constant. The same
should also apply in R.
.

 

[Antoninus Twink]

Is this tantamount to saying that the definition of "analytic" has
changed (in current undergraduate texts) in the last 20 years?

I don't think so, though I'm no historian. I think it's now more common
for (single-variable) complex analysis texts to use the term
"holomorphic" in preference to "analytic", but there's never been any
doubt about what "analytic" means: for example, Rudin's "Principles of
mathematical analysis", first page of chapter 8. I have the 3rd ed from
1976, which is well over 20 years ago! I doubt it was any different in
previous editions.

 

[Antoninus Twink]

I useful side effect of this is that on any complex surface S in C,

You mean on any Riemann surface (a "complex surface" has 4 real
dimensions). Actually, you probably just mean on C.

 

[JosephKK]

I don't think so, though I'm no historian. I think it's now more common
for (single-variable) complex analysis texts to use the term
"holomorphic" in preference to "analytic", but there's never been any
doubt about what "analytic" means: for example, Rudin's "Principles of
mathematical analysis", first page of chapter 8. I have the 3rd ed from
1976, which is well over 20 years ago! I doubt it was any different in
previous editions.

I guess i should find my old texts and check.
.

 

[JosephKK]

You mean on any Riemann surface (a "complex surface" has 4 real
dimensions). Actually, you probably just mean on C.

Yep. Wow, do i ever have a bad case of math rot. And i still can
terrorize all my cow-orkers.
.

 

[Dik T. Winter]

>
> I don't think so, though I'm no historian. I think it's now more common
> for (single-variable) complex analysis texts to use the term
> "holomorphic" in preference to "analytic", but there's never been any
> doubt about what "analytic" means: for example, Rudin's "Principles of
> mathematical analysis", first page of chapter 8. I have the 3rd ed from
> 1976, which is well over 20 years ago! I doubt it was any different in
> previous editions.

Ahlfors, Complex Analysis, 1966:
"The class of analytic functions is formed by the complex functions of
a complex variable which possess a derivative wherever the function is
defined. The term holomorphic functions is used with identical meaning."

I do not think that part of the text has changed much since the first
edition of 1953.