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What is Euler's formula

  1. Jul 23, 2014 #1

    Euler's formula, [itex]e^{ix}\ =\ \cos x\ +\ i \sin x[/itex], enables the trigonometric functions to be defined without reference to geometry.


    [tex]e^{ix}\ =\ \cos x\ +\ i \sin x[/tex]

    and so cos and sin may be defined:

    [tex]\cos x\ =\ \frac{1}{2}\left(e^{ix}\ +\ e^{-ix}\right)[/tex] and [tex]\sin x\ =\ \frac{1}{2i}\left(e^{ix}\ -\ e^{-ix}\right)[/tex]


    [tex]\cos x\ =\ 1\ -\ \frac{x^2}{2} +\ \frac{x^4}{24} -\ \frac{x^6}{720}\ \dots\ = \sum_{n\,=\,0}^{\infty}\frac{(-x)^{2n}}{(2n)!}[/tex]

    [tex]\sin x\ =\ x\ -\ \frac{x^3}{6} +\ \frac{x^5}{120} -\ \frac{x^7}{5040}\ \dots\ = \sum_{n\,=\,0}^{\infty}\frac{(-x)^{2n+1}}{(2n+1)!}[/tex]

    Extended explanation

    Proof of Euler's formula, starting from the trignonometric definitions of cos and sin:

    Using the chain rule:

    [tex]\frac{d}{dx}\left(e^{-ix}\,(cosx\ +\ i sinx)\right)[/tex]

    [tex]=\ e^{ix}\,(-i cosx\ +\ sinx\ -\ sinx\ +\ i cosx)[/tex]

    [tex]=\ 0[/tex]

    and so [itex]e^{-ix}\,(cosx\ +\ i sinx)[/itex] is a constant. Setting x = 0 we find that this constant must be 1.

    and so [tex]cosx\ +\ i sinx\ =\ e^{ix}[/tex]


    Euler's formula was discovered by Cotes.

    de Moivre's formula, [itex](cosx\ +\ i sinx)^n[/itex] = [itex]cos(nx)\ +\ i sin(nx)[/itex], is an obvious consequence of Euler's formula, but was discovered earlier.

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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