
This is more of a maths question than a
science question, but I'll ask here anyway, since
I doubt it'll be covered at a High School
level: Why, precisely, is e^{(pi *
i)} + 1 = 0 ? (I'm aware that
e^{(pi*x)} = sin x + cos x, but I'm not
sure why that is the case, either.)

Question Date: 20100603   Answer 1:
The simple answer is that your second formula
is just a bit wrong: it's actually e^{(i*x)
} = cos(x) + i*sin(x). If x = pi, then cos
(pi) = 1, sin(pi) = 0, and the equation
simplifies to e^{(i*pi)} = 1, which can
be rearranged to e^{(i*pi)}  1 = 0.
The more complicated question is why e^{
(i*x)} = cos(x) + i*sin(x). This requires
some calculus.
First, we have to know the definition of the
exponential function exp(x) or e^{x}
( 'exp' is a common way of writing the
exponential function 'e^').
The definition of exp(x) is the function that
is the derivative of itself. In other words,
the value of 'e' is such that the slope of exp
(x) is always exp(x). So in calculus notation,
if F = exp(x), dF/dx = exp(x) = F. If F = exp
(i*x), then using the chain rule, dF/dx = i*exp
(x) = i*F.
OK, now suppose we have a function F = cos(x)
+ i*sin(x). The derivative of cos(x) is sin
(x), and the derivative of sin(x) is cos(x), so
dF/dx = sin(x) + i*cos(x). Now doing some
algebra we get: dF/dx = i*( cos(x)  sin(x)/i )
= i*( cos(x) + i*sin(x) ); (to do the last
step, remember that i = sqrt(1), and multiply
sin(x)/i by i/i).
So dF/dx = i*( cos(x) + i*sin(x) ) = i*F.
This means that the derivatives of both cos(x) +
i*sin(x) and exp(i*x) are i times the original
function. So the two functions are equivalent,
and e^{(i*x)} = cos(x) + i*sin(x).
That is kind of a simple proof of the
concept, if you can believe it. More formal
proofs are much more complicated (and harder to
write down in an email!).
This is a really useful equation for many
reasons, mostly because exponential functions
are much easier to use mathematically than
trigonometric functions. Also, any complex
number can be written as the sum of its Real and
Imaginary parts, z = Re + Im = x + i*y. This in
turn can be written as z = z*(cos(p) + i*sin
(p)), where p is an angle in the complex number
plane. So complex numbers can also be written
using simple exponential functions, which is
pretty useful, too. Wikipedia has a pretty good
article on this, in
complex_number
and
Euler
if you'd like to do some more reading.
  Answer 2:
Sine and cosine can be written in
mathematical expansion series as the following:
e^{x} = (sum from a = 0 to infinity)
(x^{a})/a! = 1 + x + x^{2}/2 +
x^{3}/6 + x^{4}/24 +
x^{5}/120 + ...
If you substitute i*x for x, you get alternating
positive and negative real and imaginary terms.
If you add up all of the real terms, the result
is a cosine function in which x is measured in
units called radians, of which there are 2*pi in
a complete cycle, and likewise the imaginary
terms add up to a corresponding sine function,
also in radians.
Measured in radians, cos(pi) = 1, because
it's half way (i.e. 180 degrees) around the
cycle, and similarly sin(pi) = 0. Therefore...
e^{(pi*i)} = cos(pi) + i*sin(pi) = 1 +
i*0 = 1
1 + 1 = 0
This function is actually EXTREMELY useful to
physics, because light waves manifest as
alternating electric and magnetic fields, of
which the electric field follows a cosine
function and the magnetic field follows a sine
function at the same value of theta or x, and in
which magnetism is essentially imaginary
electricity.
  Answer 3:
The best way to show this is with a Taylor
Series expansion. Any
continuous function can be expanded as an
infinite sum of polynomials,
i.e. f(x) = Sum(d^{n}/dx^{n}(f
(x))/n!x=a (xa)^{n}) over all integer
n. By
finding the Taylor Series expansions of
e^{x}, sin(x) and cos(x), one can
show
e^{(i*x)} = cos(x) + i*sin(x).
e^{(i*pi)} = cos(pi) + i*sin(pi) = 1 +
0.
Basically, you need some calculus taking
derivatives of sin, cos and e^{x}
to prove the Taylor Series expansions. Once you
know e^{(i*x)} = cos(x) +
i*sin(x), though, you can go from there!
  Answer 4:
You are right that your question goes beyond
the high school. A full explanation could
involve calculus, differential equations, and/or
complex analysis. I don’t know what level of
math you have had, but in my explanation, I will
assume that you are familiar with trig functions
(sine and cosine) and imaginary numbers.
Euler’s formula says that
e^{(/i/*x}) = cosx + /i/*sinx,
(where x is radians). If we simply accept
Euler’s formula, it is easy to see that
e^{(/i/*pi)} = 1, because cos(pi)=1
and sin(pi)=0.
Now, if you want to understand Euler’s
formula, you need to at least know some
calculus. For this part of the explanation, I
will assume that you are familiar with
derivatives.
Any function (meeting certain conditions,
which I won’t go into here) can be represented
as what is known as an infinite series known as
a Taylor series:
f(x) = f(a) + f’(a)*(xa)/1! + f’’(a)*(xa)
^{2}/2! +f’’’(a)*(xa)^{3}/3! +
… + f^{(n)} (a)*(xa)^{n}/n! + …
For the case of a=0, this is known as a
MacLaurin series.
Recall that d/dx(e^{(ax))}=a*e^{
(ax)
}.
So, e^{(/i/x)}=e^{(0)} +
[/i/*e^{(0)}*x]/1!+ [(/i/)
^{2}*e^{(0)}*x^{2}]/2!+
[(/i/)^{3}*e^(0)*x^{3}]/3!
+ [(/i/)^{4}*e(0)*x^{4}]/4! +
[(/i/)^{5}*e(0)*x^{5}]/5! +[(/i/)
^{6}*e(0)*x^{6}]/6! + …
Simplifying (using the fact that e^{(0)
}=1,
/i/^{1}=/i/,
/i/^{2}=1,
/i/^{3}=/i/,
/i/^{4}=1,
/i/^{5}=/i/,
/i/^{6}=1, etc)
we get:
e^{(/i/x)} = 1 +/i/x – x^{2}/2!
 /i/x^{3}/3! + x^{4}/4!
+ /i/x^{5}/5! – x^{6}/6! + …
Rearranging the terms:
e^{(/i/x)} = (1– x^{2}/2! +
x^{4}/4! – x^{6}/6! + …)
+ /i/(x x^{3}/3! + x^{5}/5! + …)
You can verify that the MacLaurin series for
cosx= 1– x^{2}/2! + x^{4}/4!
– x^{6}/6! + …),
and that the MacLaurin
series for sinx= x x^{3}/3! +
x^{5}/5! + …
Therefore e^{(/i/x)}=cosx + /i*/sinx,
and we have proven Euler's formula.
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