

A lot of effort to prove that e^(*pi) + 1 = 0 is ugly - Sodel
http://symbo1ics.com/blog/?p=1089

======
reso
As an avid appreciator of mathematical elegance, I actually tend to agree with
the author on this one. This equation arises mostly from the construction of
the complex plane, and doesn't actually represent a deep relationship between
our usual definition of the constants it contains.

------
ithkuil
I hoped the author would go ahead and empathize that it's also artificial to
bring out 1 and 0 as "important" constants.

After all, as author says, the formula can also be exp(i 180∘)=-1, and 180 is
(-1) - (-1) - (-1) - (-1) - (-1) .... after all. (or any other number given
the appropriate definition of trigonometrical functions itself)

The fact that we give a special meaning to the + operation and to 1, is just
psychological, it's just a matter of defining symbols.

So let's define a=-1 and the operator , as -, we get:

exp(sqrt(h) (h,h,h,h,h,h....,h))=h

So all we have is: e^x, subtraction, multiplication, -1, and square root (of
-1 again). While the original syntax for the formula has, e^x, addition,
multiplication, i, pi, 0, 1. So we reduced the number of concepts at the
expense of making the expression longer (183 repetitions the symbol h)

But if we accept author's freedom to choose an arbitrary angle measure unit,
we could define another unit for angles and have:

exp(sqrt(-1))=-1 (so only e^, sqrt and -1: magical link between e and -1)

Btw, there is also no need for operators to use 'symbols', letters are
symbols, it's just a convention (as operator precedence), so let's rephrase
it:

d'oh hahahahahah...h ugh

Isn't that beautiful? Homer Simpson is hiding in the most elegant equation!

legend:

d -> exp

' -> (

o -> sqrt

h -> -1

a -> -

u -> )

g -> =

(implicit) -> * (lower operator precedence than -)

I guess that the what's beautiful of this equation, it's not that it's short,
but it's as short as can be by putting as many constants in it and yet don't
appear to be trivially simplifiable. E.g.

given that pi = ln(-1) / sqrt(-1)

e^(sqrt(-1) * ln(-1) / sqrt(-1)) = -1

e^(ln(-1)) = -1

-1 = -1

EDIT: that's obvious just because we type it as ln(-1) instead of i * pi, but
they do have the same value, so they are interchangeable symbols.

btw, IANAM, so not sure about the math, but as computer scientist all this
sounds to me as playing with symbols, so one could argue that you can
demonstrate that ln(-1) is pi * i by using euler identity, but perhaps going
the other way around has the same mathematical value? any idea?

EDIT: I know that the ln(-1) = i * pi is a direct result of the euler
identity, so using it to prove euler's identity is bogus. I meant we could
define ln(-1) as being a primitive constant and the expression would have a
different shape. It was just an example of why I think the canonical way of
expressing the euler's identity is called beautiful, because it's not
trivially simplifiable.

