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[edit] March 7

[edit] i^i

I was just messing around on my graphing calculator when I noticed that ii = 0.207879576...

This surprised me a lot and I was wondering- where did this number come from, and how is my calculator evaluating ii? Thanks, 68.231.151.161 04:13, 7 March 2007 (UTC)

The right hand side is the (beginning of) the decimal expansion of e^{-\frac{\pi}{2}}. Exponentiation is generalized to complex numbers via the complex logarithm. Read the article for more info. You can check out this page for a derivation of the value of ii. Phils 04:44, 7 March 2007 (UTC)
The simple explanation is this: by Euler's formula, eix = cosx + isinx. If we substitute x for π/2, that reduces to just i (since cos(π/2) = 0 and sin(π/2)=1). That is i = eiπ / 2. Now lets exponentiate that with i: i^i = (e^{i\pi /2})^i = e^{i^2\pi /2} = e^{-\pi /2} Oskar 07:54, 7 March 2007 (UTC)
Taken all at once, this ii calculation is somewhat mysterious. In taking it apart, the most important concept is embodied in Euler's formula,
e^{\bold{i} \theta} = \cos \theta + \bold{i} \sin \theta . \,\!
As θ goes from 0 to 2π the exponential spins around a unit circle in the complex plane. There are ways to make this fact seem natural, but let's just take it as given. Notice that i itself is a quarter turn away from 1, corresponding to θ = π⁄2. Thus we may write
\bold{i}^{\bold{i}} = \left( e^{\bold{i} \frac{\pi}{2}} \right)^\bold{i} . \,\!
Now we employ a familiar law of exponents, fortunately still valid for complex numbers.
\left( a^b \right)^c = a^{b c} \,\!
Since i2 = −1 by definition, we obtain
\bold{i}^{\bold{i}} = e^{- \frac{\pi}{2}} . \,\!
This is no longer a complex exponential; the result is purely real. --KSmrqT 08:46, 7 March 2007 (UTC)
Be careful because ez is not one-to-one (i.e. logarithm is a multivalued function). So the θ you get depends on what values you restrict θ to be (i.e. where you place your branch cuts for the logarithm). So sure i = e^{\frac{i\pi}{2}}, but also i = e^{\frac{5i\pi}{2}} = e^{\frac{9i\pi}{2}} = e^{\frac{-3i\pi}{2}} = e^{(2k + \frac{1}{2})i\pi}, for any integer k. Depending on how you restrict your theta, you could for example also say that i^i = e^{- \frac{5\pi}{2}} or i^i = e^{\frac{3\pi}{2}}; there are infinite possibilities. --Spoon! 10:51, 7 March 2007 (UTC)
Spoon is definitely right. You should understand that the power function in complex numbers is defined in terms of the complex logarithm function (also noted above), and that we define
zw = ewlogz,
usually taking the principal branch of the logarithm (so that we don't get the messy multivalued business Spoon was mentioning above). Functions of a complex variable are weird, so be careful when dealing with things that look familiar (because it is entirely possible that your previous notions of certain functions might be flawed). For instance, did you know that f(z) = ez is periodic, and unbounded? –King Bee (τ • γ) 15:45, 7 March 2007 (UTC)
Of course I fully endorse these cautions about multiple angles; hence my explicit restriction, [0,2π). Even so, let's not get too carried away; even real inverse functions, such as √x, can already be multiple-valued. For a naive inquiry, as this appears to be, we might not want to dive deeply into Riemann surfaces, branch cuts, and all that. When the calculator reports that the square root of 5 is approximately 2.2360679775, we usually don't fret that its negative was overlooked (even though sometimes we must). The question here was about how the calculator came up with its answer, not what other answers might be possible. --KSmrqT 22:27, 7 March 2007 (UTC)

[edit] Series

I'm trying to find out how to express different sums using binomials : Obviously, \sum_{i=1}^n{i} = \frac{n(n+1)}{2} = \binom{n+1}{2}. I then tried to find out for sums of i2 : knowing that every square is the sum of two consecutive triangular numbers and that the sum of the triangular numbers up to n is the nth tetrahedral number, I came up with \sum_{i=1}^n{i^2} = 2 \binom{n+2}{3} - \binom{n+1}{2}. I am trying to do the same for i3 but I am stuck : I tried using about the same method but didn't get anywhere (I know that \sum_{i=1}^n{i^3} = \left ( \sum_{i=1}^n{i} \right )^2, but well...)

Is there a way to express the sum of in with binomials like I did for the first two cases ? Otherwise, is there a formula which would give the sum of in for all positive integer n ?

--Xedi 15:05, 7 March 2007 (UTC)

You're looking for the Bernoulli numbers. Fredrik Johansson 15:24, 7 March 2007 (UTC)
Wow, thanks a lot, perfect ! --Xedi 15:27, 7 March 2007 (UTC)