What is the Taylor series for i^i?

Khan · Oct 27, 2003

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?

Njorl · Oct 27, 2003

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 · Oct 27, 2003

Hello, Khan!

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

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

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

What is the Taylor series for i^i?

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What is the Taylor series for i^i?

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