Surely, $\frac{d}{dx}e^{ax}$ is $ae^{ax}$ when $a$ is a real number. However, it does not mean $\frac{d}{dx}e^{ax}$ is $ae^{ax}$ when $a$ is a complex number, I think. if he didn’t prove it, Euler’s formula is not a formula but a definition of $e^{ix}$. Please tell me whether Euler’s formula is a formula or a definition. Additionally, I want to know whether Euler thought this equation as a formula or a definition when he found this.
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1If $x$ is real then the proof that $\frac{\mathrm d}{\mathrm dx}\mathrm e^{ax} = a\mathrm e^{ax}$ is valid, even when $a \in \mathbb C$. It's only when $x$ is allowed to be complex that things change, e.g. Cauchy-Riemann equations. – Fly by Night Nov 24 '17 at 16:04
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He used his own formula, I guess, and applied the usual derivatives of trigonometric functions:
$$\forall\,x\in\Bbb R\;,\;\;e^{ix}=\cos x+i\sin x\implies\left(e^{ix}\right)'=-\sin x+i\cos x=i\left(\cos x+i\sin x\right)=ie^{ix}$$
DonAntonio
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Your question depends on the definition of symbol $e^{ix} $. One approach is to use the power series for $e^{z} $. Using this one can easily prove the derivative formula in question. Another approach is to use the definition $$e^{ix} =\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^{n}$$ and then establish that $e^{ix} =\cos x+i\sin x$ so that the derivative formula is almost obvious. A third and more trivial approach is to define the symbol $e^{ix} $ by the expression $\cos x+i\sin x$ and then like before the result is obvious.
My guess is that Euler used the infinite series for $e^{z} $ as it is simpler but not trivial. But he might have used the Taylor series for $\sin, \cos $ to arrive at the formula $\cos x +i\sin x=e^{ix} $ which then leads us to the derivative formula.
Paramanand Singh
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