ProofOfEuler'sFormula
Forfatter:
Adrian D'Costa
Sidst opdateret:
7 år siden
Licens:
Creative Commons CC BY 4.0
Resumé:
Proof of Euler's Formula
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
Proof of Euler's Formula
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
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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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