On Thu, Feb 08, 2001, Oleg Goldshmidt wrote about "Re: [hackers-il] [Fwd: The Unreasonable Effectiveness of Mathematics]":

> Moshe Zadka <moshez@...> writes:

>

> > The usual definition is

> > oo

> > e^x = sigma x^n*(1/n!)

> > n=0

>

> Actually, the "usual" definition is lim (1+1/n)^(x*n) as n->oo ;-).

Don't laugh, but many times teachers of Infi 1 indeed define "e" in such

a way. Why? Think about it: we want to define e as the base such that

f(x)=e^x gives f'=f (or equivalently, such that log_e'(x)=1/x).

It's easy to define such an f with a taylor series (see more on that below),

but you learn about e^x before you learn Taylor series! Another way to

define e^x, by defining log_e as integral of 1/x, is also impossible before

you learn what an integral is (and in the Technion math department, you don't

learn integration until the second semester). So, you make definitions like

the one you showed above, and go ahead proving, using the basic definitions

and epsilons and deltas, that (e^x)'=(e^x) or (log_e)=1/x. I don't remember

all the details, but you may find them in some Infi books (though not all

of them go this way). I think (but this was 10 years ago, so I might be

wrong) that this is they way we were taught e^x.

But such a defintion cannot be extended to define e^z where z is complex,

because you'd still need to define how to raise a number to a complex power,

so it's a chicken-and-egg problem. Once you define a Taylor series, and

show e^x's Taylor series, you notice that it's very easy to extend the

taylor series to a wider selection of x's: you can substitute complex z's

instead of the x's, and can even substitute other crazy things, such as

whole matrices, and obtain a defintion for e^A, where A is a matrix.

Anybody who learned ODE (ordinary differential equations) will probably

know this fact.

By the way, if you think that defining the complex e^z by its Taylor series

is strange, awkward, or ad-hoc, then try thinking for a moment why e^x is

used at all, and not, say, 10^x or 2^x? The answer is that e^x is the only

base that results in a function f=f', and that (together with e^0=1)

immediately results in e^x's familiar Taylor series. So, if you want a

function f(z) whose f'(z)=f(z) even for complex numbers, then you'll need

to define it using that same taylor series.

The e^A for matrices I mentioned above is useful for the same reason:

(e^A)'=e^A (where derivative is defined as expected), and that is true

exactly because of the taylor series used to define it.

> Of course, you are right that exp can be defined through the

> series, but a lot of logical steps become hidden, such as analyticity

> of the function, which guarantees that the Taylor expansion is unique,

> and thus suitable for defining exp.

I don't remember the details (which is, frankly, quite shameful...), this

logic seems a bit circular. You can't say e^z is "inherently analytic"

before you actually define it. What you can do is say that e^x must have

a unique analytic extension, and that the taylor series is indeed one of

them (1. analytic and 2. an extension) and thus the only possible one.

But I don't remmeber the details, or the order in which teachers usually

build the definition of e^z. Correct me if I'm wrong - it has been many

years since I last touched this subject :(

--

Nadav Har'El | Thursday, Feb 8 2001, 16 Shevat 5761

nyh@... |-----------------------------------------

Phone: +972-53-245868, ICQ 13349191 |If I were two-faced, would I be wearing

http://nadav.harel.org.il |this one?.... Abraham Lincoln