Taylor series for i^i?

  • Thread starter Khan
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
Njorl
Science Advisor
258
10
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
 
  • #3
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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