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December 6th, 2018, 07:54 AM   #1
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transcendental equation

How to solve it ?
$\displaystyle x=(-1)^x$
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December 6th, 2018, 10:47 AM   #2
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-1 = (-1)^(-1) .... I think!!!
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December 6th, 2018, 04:50 PM   #3
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Does $x$ have to be real?
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December 6th, 2018, 06:43 PM   #4
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Quote:
Originally Posted by skipjack View Post
Does $x$ have to be real?
Reminds me of that great old song, "You got to be real!"

Quote:
Originally Posted by Denis View Post
-1 = (-1)^(-1) .... I think!!!
That's what I get.

$-1^{-1} = e^{-1 \log(-1)}

= e^{- i \pi} = \frac{1}{e^{i \pi}}

= \frac{1}{-1} = -1$.

I didn't bother to consider the alternate values of $\log i \pi$ but I'll leave that to the reader.
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Last edited by Maschke; December 6th, 2018 at 06:48 PM.
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December 6th, 2018, 10:39 PM   #5
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if not


$\left(
\begin{array}{c}
-2.96279-0.347906 i \\
-4.96731-0.511894 i \\
-6.9718-0.619368 i \\
-8.97524-0.699485 i \\
-10.9779-0.763401 i \\
-12.98-0.816587 i \\
-14.9817-0.862137 i \\
-16.9831-0.901972 i \\
-18.9843-0.937368 i \\
-20.9853-0.969218 i \\
\end{array}
\right)
$

are a bunch of solutions. Mathematica calls it

$\dfrac{i W_{k}(-i \pi )}{\pi }$

$W$ is the Lambert W function
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December 7th, 2018, 01:28 PM   #6
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Did the W-function find $\displaystyle x=-1$ ?(without complex x)

I dont know how to solve it but i see that solution is $\displaystyle x=-1$
What about applying more things if we show at least one equivalence
$\displaystyle x=(-1)^x \; \leftrightarrow \; x(-1)^x =1$
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