|
|
|
Viewing Version
9
of
'Euler relation'
|
[ view 'Euler relation'
|
back to history
]
| Title of object: |
Euler relation |
| Canonical Name: |
EulerRelation |
| Type: |
Definition |
| Created on: |
2001-11-08 07:34:17 |
| Modified on: |
2005-06-02 17:39:48 |
| Classification: |
msc:30B10 |
| Synonyms: |
Euler relation=Euler's formula
Euler relation=Euler identity |
Revision comment (for changes between this and next version):
| Changes for correction #12694 ('Other names for Euler's relation'), also other changes for style and clarity. |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{xypic} |
Content:
\PMlinkescapeword{fields}
\PMlinkescapeword{terms}
{\em Euler's relation} (also known as {\em Euler's formula}) is considered the first \PMlinkescapetext{bridge} between the fields of algebra and geometry, as it relates the exponential function to the trigonometric sine and cosine functions.
The goal is to prove
\[
e^{ix} = \cos{x}+i\sin{x}
\]
for any complex number $x$.
It's easy to show that
\begin{eqnarray*}
i^{4n} & = & 1,\\
i^{4n+1} & = & i,\\
i^{4n+2} & = & -1,\\
i^{4n+3} & = & -i
\end{eqnarray*}
for\, $n = 0,\,1,\,2,\,...$
Now, using the Taylor series expansions of $\sin x$, $\cos x$ and $e^x$, we can show that
\begin{eqnarray*}
e^{ix} & = & \sum_{n=0}^{\infty} \frac{i^n x^n}{n!}\\
e^{ix} & = & \sum_{n=0}^{\infty}(\frac{x^{4n}}{(4n)!}+
\frac{ix^{4n+1}}{(4n+1)!}
-\frac{x^{4n+2}}{(4n+2)!}-\frac{ix^{4n+3}}{(4n+3)!}).
\end{eqnarray*}
Because the series expansion above is absolutely convergent for all $x$,
we can rearrange the terms of the series as
$$e^{ix} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}(-1)^n+
i\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}(-1)^n.$$
According the entry complex sine and cosine we obtain the asserted equation
$$e^{ix} = \cos{x}+i\sin{x}.$$ |
|
|
|
|
|
