- #1
fedaykin
- 138
- 3
Why is i^i = 0.2078 ?
l'Hôpital said:[tex]
e^{i\pi} = -1
[/tex]
[tex]
\sqrt{( e^{i\pi} )} = e^{i \pi/2} = i
[/tex]
Now raise to the i.
[tex]
(e^{i \pi/2})^i = e^{i^2 \pi/2} = e^{-\pi/2} = i^i
[/tex]
l'Hôpital said:[tex]
e^{i\pi} = -1
[/tex]
[tex]
\sqrt{( e^{i\pi} )} = e^{i \pi/2} = i
[/tex]
Now raise to the i.
[tex]
(e^{i \pi/2})^i = e^{i^2 \pi/2} = e^{-\pi/2} = i^i
[/tex]
Too bad..Char. Limit said:I must admit, that proof is one of the best I've ever seen of what [tex]i^i[/tex] equals.
It is not an equation, but an equality.(And [tex]e^{i\pi}+1=0[/tex] is my personal favorite math equation, so I like that you used it.)
arildno said:Too bad..
Here we go:
[tex]i=e^{(2n+\frac{1}{2})\pi{i})},n\in\mathcal{Z}\to{i}^{i}=e^{-(2n+\frac{1}{2})\pi}[/tex]
Infinite values, that is..
arildno said:It is not an equation, but an equality.
Euler's Identity - Wikipedia said:Euler's identity is also sometimes called Euler's equation.
Merriam-Webster said:Equation: a usually formal statement of the equality or equivalence of mathematical or logical expressions.
Char. Limit said:I believe, all equalities are equations.
The value of i^i is approximately 0.2078.
i^i is solved using complex numbers and the natural logarithm function.
The value of i^i is not a whole number because it involves imaginary numbers, which do not follow the rules of real numbers.
Yes, i^i can be simplified using Euler's formula e^(i*pi) = -1, which leads to the simplified expression of i^i = e^(-pi/2).
Solving i^i has implications in various fields such as physics, engineering, and mathematics. It is also an important concept in complex analysis and helps understand the behavior of imaginary numbers.