ProofOfEuler'sFormula
Author:
Adrian D'Costa
Last Updated:
há 7 anos
License:
Creative Commons CC BY 4.0
Abstract:
Proof of Euler's Formula
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Proof of Euler's Formula
\begin
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\begin{document}
{\center{\Huge Proof of Euler's Formula \par}}
{\center{\Large Adrian D'Costa \par}}
\text{ }
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We have to prove that:
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$\displaystyle e^{i\theta} = \cos{(\theta)} + i\sin{(\theta)}$
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Let $\displaystyle f(\theta) = e^{i\theta}$ According to Maclaurin series we know that:
$\displaystyle f(x) = \mathlarger{\mathlarger{\sum}}\limits_{n = 0}^{\infty}\frac{f^{(n)}(a)x^n}{n!}\,\, \text{ where } f^{(0)}(a) = f(a) \text{ and } 0! = 1$
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Also:
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$i^{2} = -1,\, i^{3} = -i,\, i^{4} = 1,\, i^{5} = i,\, i^{6} = -1, i^{7} = -i,\, i^{8} = 1$
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\begin{align*}f^{(0)}(0) &= 1\\ f^{(1)}(0) &= i\\ f^{(2)}(0) &= -1\\f^{(3)}(0) &= -i\\ f^{(4)}(0) &= 1\\ f^{(5)}(0) &= i \\ f^{(6)}(0) &= -1\\f^{(7)}(0) &= -i\\ f^{(8)}(0) &= 1\end{align*}
$\displaystyle e^{i\theta} = \frac{1}{0!} + \frac{i\theta}{1!} - \frac{\theta^{2}}{2!} - \frac{i\theta^{3}}{3!} + \frac{\theta^{4}}{4!} + \frac{i\theta^{5}}{5!} - \frac{\theta^{6}}{6!} - \frac{i\theta^{7}}{7!} + \frac{\theta^{8}}{8!} + ...$
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Now to find Maclaurin series representation of:
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$\displaystyle g(\theta) = \cos{(\theta)}$
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\begin{align*}g^{(0)}(0) &= \cos{(0)} = 1\\ g^{(1)}(0) &= -\sin{(0)} = 0 \\ g^{(2)}(0) &= -\cos{(0)} = -1 \\ g^{(3)}(0) &= \sin{(0)} = 0 \\ g^{(4)}(0) &= \cos{(0)} = 1 \\ g^{(5)}(0) &= -\sin{(0)} = 0 \\ g^{(6)}(0) &= -\cos{(0)} = -1 \\ g^{(7)}(0) &= \sin{(0)} = 0 \\g^{(8)}(0) &= \cos{(0)} = 1 \end{align*}
$\displaystyle \cos{(\theta)} = \frac{1}{0!} - \frac{\theta^{2}}{2!} + \frac{\theta^{4}}{4!} - \frac{\theta^{6}}{6!} + \frac{\theta^{8}}{8!} + ...$
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Maclaurin series representation of:
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$\displaystyle h(\theta) = \sin{(x)}$
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\begin{align*}h^{(0)}(0) &= \sin{(0)} = 0\\ h^{(1)}(0) &= \cos{(0)} = 1 \\ h^{(2)}(0) &= -\sin{(0)} = 0 \\ h^{(3)}(0) &= -\cos{(0)} = -1 \\ h^{(4)}(0) &= \sin{(0)} = 0 \end{align*}
\begin{align*} h^{(5)}(0) &= \cos{(0)} = 1 \\ h^{(6)}(0) &= -\sin{(0)} = 0 \\ h^{(7)}(0) &= -\cos{(0)} = -1 \\h^{(8)}(0) &= \sin{(0)} = 0 \end{align*}
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$\displaystyle i\sin{(\theta)} = \frac{i\theta}{1!} - \frac{i\theta^{3}}{3!} + \frac{i\theta^{5}}{5!} - \frac{i\theta^{7}}{7!} + ...$
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Now:
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\begin{align*}\cos{(\theta)} + i \sin{(\theta)} &= \frac{1}{0!} - \frac{\theta^{2}}{2!} + \frac{\theta^{4}}{4!} - \frac{\theta^{6}}{6!} + \frac{\theta^{8}}{8!} + \frac{i\theta}{1!} - \frac{i\theta^{3}}{3!} + \frac{i\theta^{5}}{5!} - \frac{i\theta^{7}}{7!} + ... \\ &= \frac{1}{0!} + \frac{i\theta}{1!} - \frac{\theta^{2}}{2!} - \frac{i\theta^{3}}{3!} + \frac{\theta^{4}}{4!} + \frac{i\theta^{5}}{5!} - \frac{\theta^{6}}{6!} - \frac{i\theta^{7}}{7!} + \frac{\theta^{8}}{8!} + ... \\ &= e^{i\theta} \end{align*}
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\text{[Q.E.D.]}
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\section{Another way to prove Euler's formula}
Another way to prove Euler's formula is that:
$\displaystyle \cos{(\theta)} + i \sin{(\theta)} = e^{i\theta}$
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Now rearranging that equation we get:
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$\displaystyle 1 = \frac{e^{i\theta}}{\cos{(\theta)} + i \sin{(\theta)}}$
\begin{align*}0 &= \frac{d}{d\theta}\left(\frac{e^{i\theta}}{\cos{(\theta)} + i\sin{(\theta)}}\right).....\text{[Taking der. of both sides]} \\0 &= \frac{ie^{i\theta}\cos{(\theta)} - e^{i\theta}\sin{(\theta)} + e^{i\theta}\sin{(\theta)} - ie^{i\theta}\cos{(\theta)}}{\cos^{2}{(\theta)} + 2 i \cos{(\theta)}\sin{(\theta)} - \sin^{2}{(\theta)}}\\ 0 &= \frac{0}{\cos^{2}{(\theta)} + 2 i \cos{(\theta)}\sin{(\theta)} - \sin^{2}{(\theta)}}\\ 0 &= 0 \\ L.H.S &= R.H.S \end{align*}
\text{[Q.E.D.]}
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