What is the Taylor series for i^i?

In summary, the conversation discusses the expansion of i^i and the use of DeMoivre's Theorem to find the value. It is noted that i^i is a constant and using DeMoivre's Theorem with x = pi/2, the value is found to be 0.207879576.
  • #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?
 
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  • #2
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
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...
 

1. What is a Taylor series for i^i?

A Taylor series for i^i is a mathematical series that represents the value of i^i, where i is the imaginary unit.

2. How is a Taylor series for i^i derived?

A Taylor series for i^i is derived by using the Taylor series expansion formula, which calculates the value of a function at a given point by using the function's derivatives at that point.

3. What is the first few terms of the Taylor series for i^i?

The first few terms of the Taylor series for i^i are: i^i = 1 + i(ln(i)) - 1/2(ln(i))^2 - 1/6(i)(ln(i))^3 + 1/24(i)(ln(i))^4 + ...

4. How accurate is the Taylor series for i^i?

The accuracy of the Taylor series for i^i depends on the number of terms used. The more terms used, the closer the approximation will be to the actual value of i^i.

5. What is the practical use of the Taylor series for i^i?

The Taylor series for i^i is used in various fields of mathematics and physics, such as complex analysis and quantum mechanics, to approximate the value of i^i and make calculations more manageable.

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