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

On Tue, 17 Mar 2009 19:29:06 +0000 (UTC), Kaz Kylheku
<kkylheku@xxxxxxxxx> wrote:

On 2009-03-17, Antoninus Twink <nospam@xxxxxxxxxxxxxx> wrote:
On 17 Mar 2009 at 6:08, Nate Eldredge wrote:
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).

Note that c can be zero, and so the function f(x) = 0 is also its own
derivative.

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.

Neat! The only solution that came to my mind was invoking uniqueness
results for solutions of linear first order ODEs, and I was sure that
was overkill.

Neat? Nate's proof smells awfully vacuous. I suspect it relies on the assumed
properties of exp(x) to prove something about exp(x).

Suppose that you approach the proof without any prior knowledge about exp(x)
being e to the power of x, and all of the properties which that entails.

What are the weakest properties of this mysterious exp(x) for the proof's
argument to be valid?

The proof asks us to accept that whatever exp(x) is, it differentiates to
itself. That is to say exp(x) denotes any member of a class of functions that
differentiate to themselves. This allows the product rule step to to hold. So
far so good. It's valid to imagine a set of self-differentiating functions,
give it a name, and plug it into a formula where we apply a valid reduction
from differential calculus. So far so good! Of course g(x) is constant.

But then, we are asked to accept that this exp(x) function has the property
that exp(x) * exp(-x) is a constant, which is what allows us to algebraically
identify f(x) as c * exp(x).

Hence the proof's conclusion is bootstrapped from its own assumptions. It
/assumes/ that all functions exp which self-differentiate have the property
that exp(x) = c/exp(-x), and since f(x) is a self-differentiating function,
it must be one of these.

It's always good to think about these things but your argument is
not to the point. It is trivial to show that if there is a
function exp'(x) = exp(x) then exp(x)*exp(-x) = 1. Just
differentiate both sides. However that fact is not needed in
Nate's proof. It works because d(exp(-x))/dx = -exp(-x) from the
chain rule.




Richard Harter, cri@xxxxxxxx
http://home.tiac.net/~cri, http://www.varinoma.com
If I do not see as far as others, it is because
I stand in the footprints of giants.
.

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