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

"Dik T. Winter" <Dik.Winter@xxxxxx> writes:

In article <slrngrt7db.fu1.nospam@xxxxxxxxxxxxxx> Antoninus Twink <nospam@xxxxxxxxxxxxxx> writes:
> On 16 Mar 2009 at 12:41, Dik T. Winter wrote:
> > In article <slrngrqvit.12u.nospam@xxxxxxxxxxxxxx> Antoninus Twink <nospam@xxxxxxxxxxxxxx> writes:
> > > Is it obvious that there is a unique function on R that is its own
> > > derivative? Is it obvious that there aren't non-analytic functions
> > > with that property?
> >
> > It must be immediately clear there are no non-analytic functions with
> > that property. Analytic means that the function has a derivative in
> > the region over which it is analytic.
>
> Perhaps I am being dim, but why does that immediately imply it?

The definition of analytic is that it has a derivative over its domain.
So if the requirement is that f'(x) = f(x) over the domain, the function
f can not be non-analytic because if it non-analytic over the domain there
would be an x for wich the derivative does not exist.

That's not the definition of "analytic" that's usually used in
mathematics. Antonius has given the correct definition: a function f is
analytic at a point c if there is a power series

a_0 + a_1 (x-c) + a_2 (x-c)^2 + a_3 (x-c)^3 + ...

which converges to f(x) for all x in some neighborhood of c. f is
analytic if it is analytic at every point in R. (We often use the term
"real analytic" when talking about functions on R; "analytic" by itself
usually refers to functions on C.)

It's a theorem that if f is analytic, then it is infinitely
differentiable (has derivatives of all orders), and the power series
which converges to f is actually the Taylor series of f. In the case of
functions on R, the converse of this statement is false. See below.

> By analytic, I mean given by a convergent power series. For functions
> from C to C, this is equivalent to being infinitely differentiable (in
> the complex sense), but for functions from R to R, being analytic is a
> stronger property.

Is it? What more is required?

Yes, it is.

There are functions on R which are infinitely differentiable but for
which the power series at some point does not converge to the function
in a neighborhood of that point.

The classical example is

f(x) = exp(-1/x^2) for x > 0
f(x) = 0 for x <= 0

You can verify that f has derivatives of all orders at every point, and
in particular at x=0. Moreover, at x=0, the derivatives of all orders
are equal to 0. So if you expanded f in a Taylor series about x=0, you
would get the series 0 + 0 + 0 + 0.... The power series converges to 0
at every point, but that's not the value of f(x) for any x > 0.

So f is a function which is infinitely differentiable but not real
analytic.

The point is that any argument which implicitly assumes that the
function can be written as a power series will only apply to analytic
functions.

> I happily agree that it's immediately clear that
> exp(x) = 1 + x + x^2/2! +...
> is the unique analytic function from R to R (or from C to C) with
> exp'=exp, but I don't see a simple reason why there couldn't be a
> non-analytic function f:R->R with f'=f (of course, f could not be the
> restriction of a holomorphic function on C).

Can you show a non-analytic function in your sense where the derivative
does not exist for some 'x' within the domain?

Er, that's easy: f(x) = |x| is not analytic and not differentiable at
x=0. With more work I can give you one that's not analytic and not
differentiable at any point. Was that really what you meant to ask?

Anyway, exp(x) is not the only function on R which is its own
derivative; c*exp(x) for any real constant c will also do. But those
are the only examples. Here's a simple proof that doesn't assume f is
analytic.

Theorem. Suppose that f is a once-differentiable function on R such
that f'(x) = f(x) for all x in R. Then there exists a constant c such
that f(x) = c * exp(x).

Proof. Let f be such a function, and let g(x) = f(x) * exp(-x). By the
product rule, g'(x) = f'(x) * exp(-x) - f(x) * exp(-x). Since f'(x) = f(x),
g'(x) = 0 for all x. By the mean value theorem, g is constant. QED.

HTH, etc.
.

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