

(e^(i * pi)) + 1 = 0 - shawndumas
http://en.wikipedia.org/wiki/Eulers_identity

======
endtime
I remember idly putting e^(i*pi) into my TI-8x in high school when I was
bored, entirely at random, and assuming I'd broken my calculator when it
returned -1.

~~~
alan-crowe
If you like that kind of surprise you might wish to revisit cos^2 x + sin^2 x
= 1. Try it with x a complex number. Why on earth does that work? It is not as
though a complex number makes sense as an angle. More at
[http://lydianrain.wordpress.com/2009/01/06/summoning-
cthulhu...](http://lydianrain.wordpress.com/2009/01/06/summoning-cthulhu-
sin2-cos2-with-power-series/)

You could also look at Chebyshev polynomials

T_0(x) = 1

T_1(x) = x

T_2(x) = 2x^2 - 1

T_3(x) = 4x^3 -3x

It is easy enough to find recurrance relations and plots on the web. Notice
that all the graphs go from -1 to 1 and illustrate that the polynomials are
strangely well behaved in this range, oscillating up and down between -1 and
1. Why is that?

T_n(x) = cos (n arccos x)

There is this nice trignometric formula, but since cosine is always between -1
and 1 it makes no sense to try using its inverse function, arccos, outside
this range. On the other hand the polynomials have no implied restriction on
x, so if you have a calculator, such as a Common Lisp REPL, with full support
for complex numbers you can use the trignometric formula outside the -1 to 1
range and observe it agreeing with the polynomials.

------
tingley
In high school, I remember learned this from reading James Gleick's book about
Feynman. Feynman had written in one of his notebooks, as a fifteen year-old:
THE MOST REMARKABLE FORMULA IN MATH: e^(i*pi) + 1 = 0

I marched in to my calculus teacher the next day and stammered about how
unbelievable this was. A few weeks later he helped us work through why it's
true.

(Edit: ha, typo. Thanks BoppreH.)

~~~
BoppreH
e^(i*pi) + 1 = 0

It's a plus there.

------
SilentCal
There's actually a reasonably intuitive reason this one works. (At least for
me, the Taylor expansions just look like a crazy coincidence). If you try to
think about complex powers using the normal definition of exponentiation,
you'll just get a headache. Multiply e by itself i times?

However, there's another way to think about exponentiation. An exponential
function's derivative with respect to the exponent is proportional to the
value of the function, with the base determining the constant of
proportionality. This fact can be intuited. Consider the following non-
rigorous observation: x^(n+1) = x * (x^n), so x^(n+1) - (x^n) = (x-1) (x^n).

Now think about y = e^x, where d/dx (e^x) = (e^x). In words, the growth rate
of y is the current value of y. Start with e^0=1, so the growth rate is 1. If
we increase x by dx, we increase y by dx. If we increase x by idx, we increase
y by idx. So, if we draw a vector from the origin to the position of y on the
complex plane, the vector will be in the real direction, and y will be
increasing in the complex direction, perpendicular to said vector. So the
absolute value of y isn't increasing.

Now, in general, whatever the value of y, an increase of x by idx will
increase y by iydx. Multiplication by i has the effect of a ninety-degree
rotation to the left in the complex plane; try some examples if you don't
accept this. So, whatever the current value of y, a small increase in x will
move y perpendicular to the vector pointing to y. This results in a circular
motion on the complex plane, as long as the i component of x is increased.
When y gets halfway through this circle, it's back on the real axis, at -1.
When does it get there? Well, since y has moved in a circle of radius 1, it
has traveled a distance (not displacement) of magnitude pi to get through a
semicircle. Since |dy/dx| started as one, and hasn't changed, that means the
absolute value of x has increased by pi. Since we have only increased the
imaginary component of x, then x must be i * pi. So y = -1 when x = i * pi.
e^(i * pi)= -1; e^(i * pi) +1 = 0.

------
hga
According to Howard Eves in _Foundations and Fundamental Concepts of
Mathematics_ (pages 176-7), this and much of Euler's work was achieved purely
by formal manipulation. He had great intuition, although that sometimes led
him to "absurdities" in e.g. limits.

------
shawndumas
I realize that most of you probably knew this already but I didn't and it is
staggering...

Also cool is that Google 'knows':

[http://www.google.com/search?aq=f&q=(e^(i+*+pi))+%2B+1&#...</a>

~~~
ramchip
Why would it not? The Google calculator supports complex numbers in general.

[http://www.google.com/search?hl=en&source=hp&q=%28e%...](http://www.google.com/search?hl=en&source=hp&q=%28e%5E%281%2Bi%2Api%29%29%2B1%2B3i&btnG=Google+Search)

~~~
shawndumas
Dude, I just learned what i was a month ago. I am a product of public school.
I am trying; be nice.

~~~
ramchip
Haha, but don't take it badly. I was just mentioning that you can use it for
other equations, since you mentioned it like a pre-programmed special case.

~~~
zackattack
I think he was referring to the (unnecessary) "why would it not?".

------
vladev
Interestingly - this equation contains the 5 most important constant in maths.

~~~
sesqu
And the three most common operators.

~~~
billpg
Those three operators are the first three hyper operators. 1st (Addition), 2nd
(Multiplication) and 3rd (Exponentiation).

I wonder if throwing in the 4th hyper operator, tetration, and the value phi
produces anything.

~~~
techiferous
Thanks for introducing me to tetration; I had never heard of that before.

<http://en.wikipedia.org/wiki/Tetration>

------
est
If pi == 6.28 [1] then we could have a more elegant version

e^(i*pi) == 1

[1]: <http://www.math.utah.edu/~palais/pi.pdf>,
<http://www.math.utah.edu/~palais/pi.html>

~~~
drx
Part of the beauty in the original equation is that it contains every object
thought to be elemental in arithmetic and moreover, it contains every one of
them exactly once.

It contains 0, 1, +, _, ^, which are sufficient to express all of elementary
natural arithmetics (unless I'm mistaken) -- all natural numbers can be
expressed by adding enough ones to a zero (0, 0+1, 0+1+1, ...), and if I
recall correctly, + and_ aren't enough to state all the truths in natural
arithmetics so ^ is needed.

There is a very interesting paper on this called "The saga of the High School
Identities" by S. Burris
([http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PREPRI...](http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PREPRINTS/saga.ps))

Moreover, it contains three very attractive (if not the most attractive)
mathematical constants we know. There is so much we don't know about their
relationship. We have no idea if any of these numbers is transcendental or
not: pi + e, pi − e, pi*e, pi/e, pi^pi, e^e, pi^e. We don't even know if half
of them are even irrational or not (therefore, we don't know if pi is just
another facet of e, or maybe an entirely different object!)

So to see all this (and more; one could say so much more) in one equation so
terse is so surreal a feeling.

~~~
est
> 0, 1, +, , ^, which are sufficient to express all of elementary natural
> arithmetics (unless I'm mistaken)

Well, personally I don't favor zeros in equation. What's the point of 1+0=1 ?

Leaving 0 on the right side looks _so_ unbalanced.

~~~
drx
The reason I subconsciously did it was that natural numbers have an inherent
structure to them. You can describe all natural numbers using only the objects
0 and succ, where succ = (+ 1).

An interesting fact is that this is sufficient to express + and _:

a + 0 = a a + (succ b) = succ (a + b)

a _ 1 = a a * (succ b) = a + a * b

One way to look at structures (like, say, natural numbers) is too look at what
_really_ defines them, as in, what is absolutely necessary and sufficient to
define them? In this case, 0 and succ is.

Note that I'm not saying you're wrong, everyone is entitled to their own
definition of beauty. Isn't that the beauty of it?

~~~
est
> I subconsciously did it was that natural numbers have an inherent structure
> to them.

I understand and concur you :)

But I don't think Euler's identity is meant to serve this purpose. Maybe some
other formula.

It's hard to un-like something once you already liked it very much. First
impression matters. How about e^(i*pi)-1==0 instead?

------
epoweripi
its got the five most beautiful constants in one formula e, pi , i , 1 and 0
and hence i chose it for my handle :)

~~~
BoppreH
I would put phi before a lot of those.

------
Jach
Probably my favorite equation, and the proofs for the formula are quite nice
too.

Perhaps interestingly, I have a friend who accepted this, but after proof upon
proof could not agree that 0.999... is exactly equal to 1. He eventually
conceded, but it was a pretty long period of time to that point and it wasn't
through proof.

~~~
sesqu
I find this easy to accept:

    
    
      lim_{n->inf} 1 - 1/10^n = 1

But I'm not quite sure if I agree that 0.999... = lim 0.999...

edit: according to Wikipedia, they're today defined to be equal. While
convenient, it's not obvious to me.

edit2: ah-ha! I just remembered the proof that had once upon a time satisfied
me: there is a number between any two distinct numbers. there isn't one
between 0.999... and 1, so they aren't distinct.

~~~
wlievens
Bonus points for realizing it yourself and editing your post accordingly!

------
awlnk
Hi everyone.I came across the equetion e^(i)pi)1=0 in the strangest place.It
was wrien in binary. Thats not the wird bit. It was in a crop circle at a
whoping 200ft
diameter.[http://www.cropcircleconnector.com/2010/wilton/wilton2010a.h...](http://www.cropcircleconnector.com/2010/wilton/wilton2010a.html)
Their must be some realy bord intelligent pranksters out their. Or are we
getting pointed in a direction to help us advance. Is their more to this
equetion than we actually know?

~~~
sesqu
In ASCII. This equation is famous, and crafting proper crop circles requires
some familiarity with computers for the GPS and line art. This crop circle,
while clever, is not particularly astounding, and is in fact proof against
extraterrestrial crop circles.

------
gxs
I remember a while back they polled google engineers and asked them what they
thought was the most beautiful equation in math. I believe this one came in
the winner.

------
wwortiz
Once you get involved with differential equations, especially applied
differential equations this is seen quite often.

------
spektom
All trigonometric identities can be easily derived from the Euler's formula.
Instead of learning the whole course called "trigonometry" at school we'd have
been just taught this formula!

------
fbu
e^ix = cos x + i sin x

It really has nothing to do with the value of i, it's just a convention. It
really just says that cos pi = -1 which is nice but that's it.

~~~
drx
It is actually far from just a convention. It has everything to do with the
Taylor series of the functions e^x, sin x and cos x.

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

If you substitute ix for x, you get:

e^{ix} = 1 + ix/1 + (ix)^2/(2!) + (ix)^3/(3!) + ... = = 1 + ix + (-x^2)/(2!) +
(-i)x^3/(3!) + ...

You can also derive the Taylor series for sin x and cos x, which are:

sin x = x - x^3/(3!) + x^5/(5!) + ...

cos x = 1 - x^2/(3!) + x^4/(4!) + ...

Therefore you can see that e^{ix} = cos x + i(sin x). Of course, I haven't
shown how to derive the Taylor series
(<http://en.wikipedia.org/wiki/Taylor_series>) for these functions, but I have
to stop somewhere.

I hope that I have shown you that this isn't just an arbitrary convention, in
fact far from it. The formula is so beautiful also because of the many
intricate relationships between all these elementary mathematical concepts,
including complex analysis, trigonometric functions, series, etc.

It should also be noted that this is just one of the many (in fact, aleph zero
many) different ways to prove this equivalence.

Edit: Some formatting corrections, sorry, new here.

~~~
drx
I can't resist not showing this clever derivation of the e Maclaurin series.

I referenced the fact that

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

Now, one of the many equivalent definitions of e is that it is the only real
number for which this holds:

d/dt (e^t) = e^t

This means that e^t is a real function which is so smooth, that no matter how
many times you differentiate it, you always get the same function (it turns
out that it is the only such real function)

Now, we know that e^0 = 1 (since x^0 = 1, where x != 0). Therefore, in its
Maclaurin series, the only term not depending on x should therefore be 1
(otherwise e^0 wouldn't be 1).

So know we know that e^x looks something like this:

e^x = 1 + (something)

Now we can ask ourselves which this question: since d/dx e^x = e^x, what must
also be in the e^x series, if 1 belongs to it? Well, whatever differentiates
to 1, so now we know that

e^x = 1 + x + (something)

(because d/dx (1 + x + something) = 1 + x + d/dx something)

Now we can again ask this question for our current form; what must we
differentiate in order to obtain 1 + x? And thus we get

e^x = 1 + x + x^2/2 + something

This way, if we write the two (equivalent) series this way:

e^x = 1 + x + x^2/2 + d/dx e^x = 1 + x + x^2/2 + x^3/(2*3) +

And we can complete it with the infinite Maclaurin series.

Now this is less formal than it should be and it would probably make the
formalists cringe, but I hope you get the idea. You can actually apply the
same principle for sin x and cos x, except in their case, they're actually
mutually derived from each other. I'll leave that as an exercise for the
reader (oh how fun it is to say that after reading this phrase countless
times)

~~~
Dove
Great. Now I'm going to spend the rest of the morning trying to figure out
whether you can really do it that way.

[Edit: Ah yes. The formalism goes like this: Consider the above series. By the
above argument, it differentiates to itself. QED. ]

~~~
tome
Ah, but how do you know that e^x is the _only_ thing that differentiates to
itself? :)

~~~
philh
You don't need to. The uniqueness part of the definition is never used in that
argument. (In fact, a*e^x also differentiates to itself, for any a; but that's
a trivial case.)

Uniqueness almost follows from that argument. It's now easy to see that exp is
the only analytic function satisfying exp' = exp and exp(0) = 1: if you have
another one, by the same argument, it has the same Maclaurin expansion, hence
is the same function.

However, I don't know how to prove uniqueness over all functions, not just
analytic ones.

------
karanbhangui
this is the crux of a lot of electrical engineering.

------
Alistra
old, lol

