Re: imaginary exponents -- I *want* to believe...

David C. Ullrich wrote:

On Wed, 25 Feb 2009 01:03:09 -0800 (PST), Bennett Haselton
<bennett@xxxxxxxxxxxxx> wrote:

Something's bugged me about imaginary exponents ever since learning in
high school that e^ix = cos(x) + isin(x). Now that I am passing on my
alleged wisdom to high school students, I'd like to make sure it is at
least mostly right.

After learning that x^n for natural numbers n means "x times itself n
times", it seemed like the natural way to extend this to negative
numbers and fractions would be to do it in a way that "gets you back"
to the natural numbers, where you already know the answer. So if x^(a
+b) = x^a*x^b for natural numbers, then x^(-1) would be defined such
that x^(-1)*x^3 = x^2, which leads you to x^(-1) = 1/x. And x^(1/2)
must be such that x^(1/2)*x^(1/2) = x^(1/2+1/2) = x, so x^(1/2) = +/-
sqrt(x).

But that strategy doesn't work for imaginary exponents, since the way
to "get back" to the integers would be through multiplication, so you
could try to start with (x^i)^i = x^(i*i) = x^(-1) -- but that doesn't
tell you what x^i is.

Meanwhile, the book shows how you can start with y = cosx + isinx and
do some implicit differentiation and end up with y = e^ix, but I
couldn't make myself *believe* it. I can follow the math that ends
with y = e^ix, but it seemed just like a neat "trick", and what I
couldn't convince myself is that someone else might find an equally
neat trick to prove that, say, e^ix = tan(x) + i*2^x or something
equally exotic. It seems much less convincing than the argument that
x^(-1) = 1/x, which seems to me to forcefully drive home the point
that you could never come up with a "trick" to show that x^(-1) equals
anything other than 1/x.

If the book "proves" that exp(ix) = cos(x) + i sin(x) without
first giving a _definition_ of exp(ix) then the book is doing
very naughty things - before you can prove anything about
whatever you need to define whatever.

There would be nothing _wrong_ in any absolute mathematical
sense in defining e^(ix) = 27 for all x. But if we did that then
the function e^(ix) would not have any of the properties that
we _want_ it to have.

Is there a cure for my skepticism? :)

Find that a huge number of things work out very nicely if
we define e^(ix) = cos(x) + i sin(x).

One that seems very compelling to me: We know
that for real x,

(*) e^x = 1 + x + x^2/2! + x^3/x! + ... .

Now suppose we decide we _want_ (*) to hold
for complex values of x as well - that would be nice,
wouldn't it? If you substitute ix for x in (*) (with
x real) you get exactly cos(x) + i sin(x).

There is an IMO Complex Made Simple-r way to see this.

The gist of the argument is. First prove that any function F with the
property that F(x+y) = F(x).F(y) is an exponential function e^(c.x)

Then prove that cos(x) + i.sin(x) has this property and therefore it
must be a function of the form e^(c.x) . At last find out that c = i

The argument below will be without any (set theoretic) embellishments,
but I believe it's possible to fill in remaining details for oneself.
I'd like to hear, though, if there are serious omissions.

Suppose that all we know there is a "neat" (everywhere differentiable)
function F such that for all x and y :

F(x+y) = F(x).F(y)

Theorem: F(x-y) = F(x)/F(y)

Proof: F(x) = F(x-y+y) = F(x-y).F(y)

Theorem: F(0) = 1

Proof: F(x-x) = F(x)/F(x)

Theorem: F'(x) = c.F(x) where c is a constant

Proof: F(x+h) - F(x) F(h) - F(0)
F'(x) = ------------- = F(x) ----------- = F(x).F'(0) ; h-> 0
h h

Theorem: F(x) = e^(c.x)

Proof:

F'(x) / F(x) = c -> integral dF/F = integral c.dx

-> ln(F) = d + c.x -> F(x) = C.e^(c.x)

But F(0) = 1 -> C = 1 .

Now consider: G(x) = cos(x) + i.sin(x) .

Theorem: G(x+y) = G(x).G(y) ; thus G is a function like F above.

Proof: [cos(x) + i.sin(x)].[cos(y) + i.sin(y)] =
[cos(x).cos(y) - sin(x).sin(y)]
+ i.[sin(x).cos(y) + cos(x).sin(y)]

But: cos(x).cos(y) - sin(x).sin(y) = cos(x+y) : according to well-known
sin(x).cos(y) + cos(x).sin(y) = sin(x+y) trigonometric formulas

Hence: [cos(x) + i.sin(x)].[cos(y) + i.sin(y)] = cos(x+y) + i.sin(x+y)

Therefore: cos(x) + i.sin(x) = e^(c.x)

It remains to show that c = i . Differentiate with respect to x :

- sin(x) + i.cos(x) = c.e^(c.x) -> for x = 0 : i = c .

Quod Erat Demonstrandum.

Han de Bruijn

.

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