Euler's Formula

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The proof Euler's formula, using Taylor series , is presented.

Referring to the tables of mathematical formulas , the Taylor series of \( e^z \) is given by
\( e^z = 1 + z + \dfrac{z^2} { 2! } + \dfrac{z^3} { 3! } + \dfrac{z^4} { 4! } + \dfrac{z^5} { 5! } + \dfrac{z^6} { 6! } + \dfrac{z^7} { 7! } ... + \dfrac{z^n}{n!} + ... \)

Let \( z = i x \), where \( x \) is a real number and \( i = \sqrt {-1} \) is the imaginary unit.
\( e^{(ix)} = 1 + {(ix)} + \dfrac{{(ix)}^2} { 2! } + \dfrac{{(ix)}^3} { 3! } + \dfrac{{(ix)}^4} { 4! } + \dfrac{{(ix)}^5} { 5! } + \dfrac{{(ix)}^6} { 6! } + \dfrac{{(ix)}^7} { 7! } ... \)

Take the \( i \) from inside the brackets, \( e^{(ix)} \) may be written as
\( e^{(ix)} = 1 + i x + i^2 \dfrac{x^2} { 2! } + i^3 \dfrac{x^3} { 3! } + i^4 \dfrac{x^4} { 4! } + i^5 \dfrac{x^5} { 5! } + i^6 \dfrac{x^6} { 6! } + i^7 \dfrac{x^7} { 7! } ... + i^n \dfrac{x^n}{n!} + ... \)

Note that
\( i^2 = - 1 \)
\( i^3 = i^2 i = - i\)
\( i^4 = i^3 i = 1 \)
\( i^5 = i^4 i = i \)
\( i^6 = i^5 i = - 1 \)
\( i^7 = i^6 i = -i \) ...
and so on.

Therefore \( e^{(ix)} \) may be written as
\( e^{(ix)} = 1 + i x - \dfrac{x^2} { 2! } - i \dfrac{ x^3} { 3! } + \dfrac{x^4} { 4! } + i \dfrac{x^5} { 5! } - \dfrac {x^6} { 6! } - i \dfrac{x^7} { 7! } ... \)

Let us group the terms making the real part together and the terms making the imaginary together and rewrite \( e^{(ix)} \) as
\( e^{(ix)} = ( 1 - \dfrac{x^2} { 2! } + \dfrac{x^4} { 4! } - \dfrac {x^6} { 6! } ...) + i ( x - \dfrac{ x^3} { 3! } + \dfrac{x^5} { 5! } - \dfrac{x^7} { 7! } ...) \qquad (I) \)

Referring to the
tables of mathematical formulas , the Taylor series of \( \sin x \) and \( \cos x \) are given by
\( \sin x = x - \dfrac{x^3}{ 3!} + \dfrac{x^5}{ 5!} - \dfrac{ x^7}{ 7!} + ... \)

\( \cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + ... \)

Note that the real part of \( e^{(ix)} \) in \( (I) \) is given by
\( 1 - \dfrac{x^2} { 2! } + \dfrac{x^4} { 4! } - \dfrac {x^6} { 6! } ...\) which is the Taylor series of \( \cos x \)

and the imaginary part of \( e^{(ix)} \) in \( (I) \) is given by
\( x - \dfrac{ x^3} { 3! } + \dfrac{x^5} { 5! } - \dfrac{x^7} { 7! } ... \) which is the Taylor series of \( \sin x \)

and hence the Euler formula
\[ e^{ix} = \cos x + i \sin x \]



More References and Links

  1. Taylor and Maclaurin Series with Examples
  2. Tables of Mathematical Formulas
  3. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
  4. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
  5. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8

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