The derivative for $(-1)^{x}$ is \begin{equation} \frac d{dx}\left[(-1)^x\right]=i\pi(-1)^{x} \end{equation} But why?
What happens with higher order derivatives?
Thanks in advance.
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1$i\pi$ is constant, correct? – abiessu Jun 26 '15 at 01:09
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Do you have a reference for your formula? Is the function you are considering a function from the real numbers to the real numbers? What is your definition of $\frac{d}{dx}$? – Thomas Jun 26 '15 at 01:20
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9The expression $(-1)^x$ is not defined in general. You can interpret it to mean $e^{ \pi x i}$, and the formula follows from that. But you could just as easily assume that it means $e^{3 \pi x i}$ which has the derivative $3 \pi ie^{3 \pi x i}$. – Jair Taylor Jun 26 '15 at 01:42
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Since $$(-1)^{x}=(e^{i\pi})^x=e^{i\pi x}$$
We have $$ \dfrac{d}{dx}\left((-1)^{x}\right)=\dfrac{d}{dx}\left(e^{i\pi x}\right)=i\pi e^{i\pi x}=i\pi(-1)^{x} $$ For higher order derivatives $$ \dfrac{d^{n}}{dx^{n}}\left((-1)^{x}\right)=(i\pi)^n(-1)^{x} $$
Eugene Zhang
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2@AkshatTripathi That's a pretty famous result in calculus, actually. Though I see why you're confused — How do you multiply something by itself an imaginary amount of times? The proof is too long to write in a comment, but trust us that the equation makes sense and is true. – Akiva Weinberger Aug 03 '15 at 06:19
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4@AkshatTripathi It's not that $e^{i \pi x} = -1$ but that $e^{i \pi} = -1$. By Euler formula, we have: $$e^{\theta i} = \cos \theta + i \sin \theta,$$ plugging in $\theta = \pi$ we get the desired result. – rubik Aug 03 '15 at 06:20
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Mistake 1: The function $(-1)^x$ is not well-defined, even if $x$ is a complex number.
Mistake 2: You cannot say $\left( e^{\pi i} \right)^x = e^{\pi i x}$. The multiplication of exponent rule does not extend to complex numbers. Indeed, otherwise you have problems like $1^{\pi} = e^{2\pi^2 i}$, which is clearly false.
Akiva Weinberger
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Nicolas Bourbaki
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Possibly relevant: Wolfram Alpha interprets $(-1)^x$ to mean $e^{\pi ix}$. – Akiva Weinberger Aug 03 '15 at 06:22
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We have that $e^{i\pi}=-1$, so $(-1)^x=e^{i\pi x}$, therefore $\frac{d}{dx}e^{i\pi x}=i\pi e^{i\pi x}=i\pi (-1)^x$.
It's easy to see that $\frac{d^n}{dx^n}(-1)^x=(i\pi)^n(-1)^x$.
Larara
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Observe the function $w=(-1)^z$ This is equivalent to $w=e^{z\ln(-1)}$ Now, recalling that the logarithm is a multivalued function, $$\ln(-1)=\ln\vert(-1)\vert +i\arg(-1)=i(\pi+2\pi n)$$We see that $(-1)^z$ Is equivalent to $w=e^{i\pi (2n+1)z}$ For all integers $n$. This means, except in special cases, that $(-1)^z$ has an infinite number of complex values. If we choose to let n be constant, and define $(-1)^z$ as the resulting single valued function, the complex derivative is equal to$$ i\pi (2n+1)e^{i\pi (2n+1)z}$$
