What is the Taylor series for i^i?

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SUMMARY

The Taylor series expansion for i^i can be derived using Euler's formula and DeMoivre's Theorem. By substituting x = π/2 into Euler's formula, we find that i = e^(iπ/2). Raising both sides to the power of i results in i^i = e^(-π/2), which evaluates to approximately 0.207879576. This confirms that i^i is indeed a real number.

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  • Understanding of Euler's formula: e^(ix) = cos(x) + i sin(x)
  • Familiarity with DeMoivre's Theorem
  • Knowledge of Taylor series expansion
  • Basic complex number operations
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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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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
 
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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