Taylor series for i^i?

Khan

I'm having some problems expanding i^i, could anyone help? I know it becomes a real number somehow, and I'm familiar with the e^(i * pi) expansion, but is the i^i done in the same way?

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Njorl

There is a well known expansion for a^x:

a^x=SUM[((alnx)^n)/(n!)]

Just replace a and x with i.

At first glance, it doesn't look real to me, but maybe the sum telescopes.

Njorl

Soroban

Hello, Khan!

I'm not sure what you mean by expanding ii,
since it is already a constant.

Using DeMoivre's Theorem (Euler's?): eix = cos x + i sin x,
when x = pi/2, we have: ei*pi/2 = cos(pi/2) + i sin(pi/2) = i

Raise both sides to the power i: ii = (ei*pi/2)i= e-pi/2 = 0.207879576...

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